For the equilibrium AB ( g ) ⇌ A ( g ) + B ( g ) , at a given...

$Σn = \dfrac{2}{3} + \dfrac{2}{3} + \dfrac{1}{3} = \dfrac{4}{3}$
${K}_{p} = \dfrac{{n}_{A} \times {n}_{B}}{{n}_{AB}} \times {\left(\dfrac{P}{Σn}\right)}$^{\Delta{n}_{g}}$<br/>$= \left[\dfrac{\dfrac{1}{3} \times \dfrac{1}{3}}{\dfrac{2}{3}}\right] \times {\left[\dfrac{P}{\dfrac{4}{3}}\right]}^{1}$<br/>$= \dfrac{1}{6} \times \dfrac{3P}{4} = \dfrac{P}{8}$<br/>So,$\dfrac{P}{{K}_{p}} = 8