Pressure necessary to obtain 50% dissociation of${\mathrm{PCl}}_5\text{ for }{\mathrm{PCl}}_5\rightleftharpoons {\mathrm{PCl}}_3+{\mathrm{Cl}}_2$at 500K will be equal to

If P is the total required pressure, then 
$$\begin{gathered} {\mathrm{P}}_{{\mathrm{PCl}}_5}=\frac{0.5}{1.5}\times \mathrm{P}=\frac{\mathrm{P}}{3}, \\ {\mathrm{P}}_{{\mathrm{PCl}}_3}=\frac{0.5}{1.5}\times \mathrm{P}=\frac{\mathrm{P}}{3}, \\ {\mathrm{P}}_{{\mathrm{Cl}}_2}=\frac{0.5}{1.5}\times \mathrm{P}=\frac{\mathrm{P}}{3}, \\ {\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{P}}_{{\mathrm{PCl}}_3}\times {\mathrm{P}}_{{\mathrm{Cl}}_2}}{{\mathrm{P}}_{{\mathrm{PCl}}_5}}=3\mathrm{P} \\ \mathrm{P}=3{\mathrm{K}}_{\mathrm{p}} \end{gathered}$$