At 600°C, K_p for the following reaction is 1 atm, $X\left(g\right)\rightleftharpoons Y\left(g\right)+Z\left(g\right)$
at equilibrium, 50% of X(g) is dissociated. What is the partial pressure (in atm) of X(g) at equilibrium?

$X\left(g\right)\rightleftharpoons Y\left(g\right)+Z\left(g\right)$  If P is total pressure at equilibrium. 
t = 0                            1                   0                      0
Equilibrium              0.5                0.5                  0.5
Partial pressure  $\frac{0.5P}{1.5}$         $\frac{0.5P}{1.5}$             $\frac{0.5P}{1.5}$
$\begin{array}{l}{K}_p=\frac{P_y.P_z}{P_x}\\ 1=\frac{\frac{P}{3}\times \frac{P}{3}}{\frac{P}{3}}\Rightarrow p=3atm\end{array}$ 
Partial pressure of $X=\frac{P}{3}=1atm$