The equilibrium concentrations in terms of mole are given for the reaction at $18 \mathrm{atm} \mathrm{P}$ mole at equation 

$\underset{3}{{\mathrm{PCl}}_5}\rightleftharpoons \underset{3}{{\mathrm{PCl}}_3}+\underset{3}{{\mathrm{Cl}}_2}$ The value of ${\mathrm{K}}_{\mathrm{p}}$ is ........

Total pressure P=18atm.
The number of moles of PCl_5​,PCl₃​ and Cl_2​ are 3, 3 and 3 respectively.
The mole fractions of PCl₅​,PCl_3​ and Cl₂​  are 0.333, 0.333 and 0.333 respectively. 

The partial pressures of PCl₅​,PCl_3​ and Cl_2​ are 0.333×18=6 atm,6 atm and 6 atm respectively.

Hence, the value of K_p​ is   Kp​=$\frac{{\mathrm{p}}_{{\mathrm{pcl}}_3}\times {\mathrm{p}}_{{\mathrm{cl}}_2}}{{\mathrm{p}}_{{\mathrm{pcl}}_5}}$=$\frac{6\times 6}{6}$ = $6$