1001111  | O
0110101  | 5
1110110  | v
1111001  | y
1011111  | _
0101011  | +
1111110  | ~
-------
5555555  <- Sum of bits in each column. Should give a truthy value.

1010000  | P
1010000  | P
1000011  | C
1000111  | G
-------
4020122  <- Should give a falsey value

Addr. | Dump                                            | #6 #5 #4 #3 #2 #1 #0
------+-------------------------------------------------+---------------------
00-0F | 53 3D 3E 5B 2E 2E 2E 53 5D 2E 6D 61 70 28 53 3D |  8 11  9 11  9  9  9
10-1F | 3E 52 2E 6D 61 70 28 28 5F 47 53 53 53 53 53 53 | 20 18 19 17 14 20 19
20-2F | 56 57 57 57 2C 56 29 3D 3E 52 5B 56 5D 2D 3D 53 | 30 24 32 25 26 30 29
30-3F | 2E 63 68 61 72 43 6F 64 65 41 74 28 29 3E 3E 56 | 41 37 37 32 34 38 36
40-4F | 26 31 29 2C 52 3D 5B 5F 3D 33 5E 33 2C 5F 2C 5F | 47 47 48 43 44 47 46
50-5F | 2C 5F 2C 5F 2C 5F 2C 5F 5D 29 26 26 21 52 2E 73 | 54 57 55 54 56 56 54
60-6D | 6F 6D 65 28 53 3D 3E 53 5E 52 5B 5F 5D 29       | 64 64 64 64 64 64 64

BXs&=?I&]

B     1 0 0 0 0 1 0
X     1 0 1 1 0 0 0
s     1 1 1 0 0 1 1
&     0 1 0 0 1 1 0
=     0 1 1 1 1 0 1
?     0 1 1 1 1 1 1
I     1 0 0 1 0 0 1
&     0 1 0 0 1 1 0
]     1 0 1 1 1 0 1

Sum   5 5 5 5 5 5 5

B      % Input string (implicit). Convert each char to its ASCII code, and
       % then to binary. This gives a binary matrix, with each char of the
       % input corresponding to a row
Xs     % Sum of each column. Gives a row vector
&=     % All pairwise equality comparisons
?      % If all are true
  I    %    Push 3
  &    %    Specify that the next function, namely implicit display, will
       %    take one input, instead of the whole stack which is the default
]      % End
       % Display (implicit)

OBUSE&889.

Char -> Hex -> Decimal -> Binary
O       0x4F   79         0b1001111
B       0x42   66         0b1000010
U       0x55   85         0b1010101
S       0x53   83         0b1010011
E       0x45   69         0b1000101
&       0x26   38         0b0100110
8       0x38   56         0b0111000
8       0x38   56         0b0111000
9       0x39   57         0b0111001
.       0x2E   46         0b0101110
                            -------
                            5555555

OBUSE&889. - Main link: string
O          - cast to ordinals
 B         - convert to binary
  U        - upend (reverses each to prepare for vectorised sum)
   S       - sum (vectorises)
    E      - all equal? (yields 1 if all bit-sums are equal and 0 if not)
      889. - 889.0
     &     - bitwise and (1 & 889.0 is 1; and 0 & 889.0 is 0)

OBUSE$*8?8

O  1 0 0 1 1 1 1
B  1 0 0 0 0 1 0
U  1 0 1 0 1 0 1
S  1 0 1 0 0 1 1
E  1 0 0 0 1 0 1
$  0 1 0 0 1 0 0
*  0 1 0 1 0 1 0
8  0 1 1 1 0 0 0
?  0 1 1 1 1 1 1
8  0 1 1 1 0 0 0
————————————————
∑  5 5 5 5 5 5 5

OBUSE$*8?8  Main link. Argument: s (string)

O           Ordinal; map all characters in s to their code points.
 B          Binary; convert each code point to base 2.
  U         Upend; reverse each binary array to right-align the digits.
       8?   If 8 is non-zero (it is):
   SE$          Sum the corresponding digits and test the the sums for equality.
            Else (never happens):
      *  8      Raise all binary digits to the eighth power.

Total@IntegerDigits[ToCharacterCode@#,2,7]~MatchQ~{"?";a_ ..}&

0000-0010:  54 6f 74 61-6c 40 49 6e-74 65 67 65-72 44 69 67  Total@In tegerDig
0000-0020:  69 74 73 5b-54 6f 43 68-61 72 61 63-74 65 72 43  its[ToCh aracterC
0000-0030:  6f 64 65 40-23 2c 32 2c-37 5d 7e 4d-61 74 63 68  ode@#,2, 7]~Match
0000-0040:  51 7e 7b 22-3f 1f 1f 1f-1f 1f 1f 1f-1f 1f 1f 1f  Q~{"?... ........
0000-0050:  1f 1f 1f 1f-1f 1a 1a 1a-1a 18 18 18-18 18 10 22  ........ ......."
0000-0058:  3b 61 5f 20-2e 2e 7d 26                          ;a_...}&

@(_)~diff(sum(de2bi(+_)))%RRPPPVVVW?????????________

15   22    6   15   10    9   13

@(_)~diff(sum(de2bi(+_)))
@(_)      % An anonymous function that take a variable _ as input
          % We use underscore, instead of a character, since it has the
          % most suitable binary represetation
              de2bi(+_)    % Convert the input string to a binary matrix
          sum(de2bi(+_))   % Take the sum of each column
     diff(sum(de2bi(+_)))  % And calculate the difference between each sum
    ~diff(sum(de2bi(+_)))  % Negate the result, meaning 0 becomes true,
                           % and everything else becomes false

@(_)!((_=sum(de2bi(+_)))-_(1))%RRRFVVVVVVVVV_____????

@(_)      % An anonymous function that take a variable _ as input
          % We use underscore, instead of a character, since it has the
          % most suitable binary represetation
    !     % Negate the result, meaning 0 becomes true, and everything else becomes false
        de2bi(+_)         % Convert the input string to a binary matrix
    sum(de2bi(+_))        % Take the sum of each column
 (_=sum(de2bi(+_)))       % Assign the result to a new variable, also called _
                          % It's not a problem that we use the same variable name, due
                          % to the order of evaluation
((_=sum(de2bi(+_)))-_(1)) % Subtract the first element of the new variable _
                          % If all elements of the new variable _ are identical, then this
                          % should give us a vector containing only zeros,
                          % otherwise, at least one element should be non-zero
!((_=sum(de2bi(+_)))-_(1))  % And finally, we negate this.

00000000  5f 2e 6d 61 70 28 43 3d  3e 28 22 30 22 2a 37 2b  |_.map(C=>("0"*7+|
00000010  2b 28 42 69 67 49 6e 74  28 43 29 74 6f 53 74 72  |+(BigInt(C)toStr|
00000020  69 6e 67 20 32 29 29 74  61 6b 65 52 69 67 68 74  |ing 2))takeRight|
00000030  20 37 20 6d 61 70 28 5f  2d 34 38 29 29 2e 74 72  | 7 map(_-48)).tr|
00000040  61 6e 73 70 6f 73 65 2e  6d 61 70 28 5f 2e 73 75  |anspose.map(_.su|
00000050  6d 29 2e 74 6f 53 65 74  2e 73 69 7a 65 3d 3d 31  |m).toSet.size==1|
00000060  2f 2f 5f 5f 5f 5f 5f 5f  5f 5f 5f 5f 5f 5f 5f 5f  |//______________|
00000070  5f 5f 5f 5f 5f 5f 5f 5f  5f 5f 5f 5f 5f 5f 5f 5f  |________________|
00000080  1f 1f 1f 1f 1e 1e 1e 1e  16 16 16 16 16 12 12 10  |................|
00000090  10 10 10 10 10                                    |.....|

[:(*/@:={.)[:+/2 #.inv 3 u:]NB.____UUUUUUUUDD

def Y(S):
    O=map(sorted,zip(*['{:07b}'.format(ord(W))for   W   in  S]))
    return  O[1:]==O[:-1]#V_____________

00000000: 64 65 66 09 59 28 53 29 3a 0a 09 4f 3d 6d 61 70  def.Y(S):..O=map
00000010: 28 73 6f 72 74 65 64 2c 7a 69 70 28 2a 5b 27 7b  (sorted,zip(*['{
00000020: 3a 30 37 62 7d 27 2e 66 6f 72 6d 61 74 28 6f 72  :07b}'.format(or
00000030: 64 28 57 29 29 66 6f 72 09 57 09 69 6e 09 53 5d  d(W))for.W.in.S]
00000040: 29 29 0a 09 72 65 74 75 72 6e 09 4f 5b 31 3a 5d  ))..return.O[1:]
00000050: 3d 3d 4f 5b 3a 2d 31 5d 23 56 5f 5f 5f 5f 5f 5f  ==O[:-1]#V______
00000060: 5f 5f 5f 5f 5f 5f 5f 16 16 16 16 16 16 16 16 16  _______.........
00000070: 16 16 14 14 10                                   .....