$$
\begin{array}{ccc}
&\ce{2NO2(g) &<=>& N2O4(g)}\\\hline
-&a&&b&\\
t=0&a+x&&b\\
t=t_0&a+x-2y&&b+y\\
\end{array}
$$

At the first equilibrium, $K_\mathrm{p}=\frac{b}{a^2}$. At $t=0$, we add more $\ce{NO2}$, and let it reach equilibrium at $t=t_0$. All of $a,b,x,y$ are gas pressure in Pascals. Since the temperature is constant, the value of $K_\mathrm{p}$ remains the same at $t=t_0$.

So, we can say $$K_\mathrm{p}=\frac{b}{a^2}=\frac{b+y}{(a+x-2y)^2}$$

Rearranging we get an equation quadratic in both $x$ and $y$:

$$4by^2-y(4xb+4ab+a^2)+2abx+bx^2=0$$

Solving it for $y$ we get the two solutions:

$$y_1 = \frac{-a \sqrt{a^2 + 8 a b + 8 b (2 b + x)} + a^2 + 4 a b + 4 b x}{8 b}$$

and $$y_2 = \frac{a \sqrt{a^2 + 8 a b + 8 b (2 b + x)} + a^2 + 4 a b + 4 b x}{8 b}$$

Now, to prove that $a+x-2y>a$ I'm stuck :(