How to calculate the percentage dissociation of phosgene using the pressure equilibrium constant?
Here goes a possible answer, I tried to describe it step by step so you can see your (and my) possible errors. The first part of your calculation was conceptually good, but you make the calculation wrong. 

As one of the comments say, $$K_p=\frac{(342-254)*(352-254)}{254}=33.95$$

So you have $K_p$. Then 

\begin{array}{lcccc}
 &amp; \ce{CO &amp;+&amp; Cl_2 &amp;&lt;=&gt; &amp;COCl_2}\\
\text{Initial pressure of substance} &amp;0 &amp;&amp; 0 &amp;&amp; A\\
\text{Final pressure of substance}   &amp; x &amp;&amp; x &amp;&amp; A-x\\~\\
\end{array}

With the condition that $x + x + A-x=760$, we have $A+x=760$.

Using our first calculation:

$$K_p=\frac{(x)*(x)}{A-x}=33.95$$
(Kp is the same at same $T$)

$$K_p=\frac{x^2}{760-2x}=33.95$$
here we get x=130.229, then:
dissociation of $A= 1-\frac{(A-x)}{A}$
we get dissociation of A=1-0793=0.207

or 20.7% (0.01% flew away)