The dissociation equilibrium of a gas AB_ 2 can be represented as,...

& & & & & & & 2 {AB }_{2} \left(\text{g}\right)⇌ 2AB \left(\text{g}\right) + {B }_{2}\left( g \right)
At equilibrium \text{2(1} - \text{x)}\text{2x}\text{x}
Total moles = \text{(2}\text{+}\text{x)}
Partial pressures
{P}_{{\text{AB}_{2}} \text{=}\dfrac{\text{2(1} - \text{x)}}{\text{2}\text{+}\text{x}} \text{p,}{P}_{\text{AB}} \text{=}\left(\dfrac{$\text{2x}$}{$\text{2}$$\text{+}$$\text{x}$}\right) \text{p,}{P}_{{B}_{2}} \text{=}\left(\dfrac{x}{$\text{2+x}$}\right) \text{p}
{K}_{p} = \dfrac{\text{p}{AB}^{2} {pB}_{2}}{( {pAB}_{2} {)}^{2}} = \dfrac{[ ($\dfrac{2 x}{2 + x}$)$\text{p}${]}^{2} [ \dfrac{x \text{p}}{2 + x} ]}{[ $\dfrac{2 ( 1 - x )}{2 + x} \text{p}$ {]}^{2}}$<br />${K}_{p} = \dfrac{4x^{3} \text{P}}{4 \left(2 + \text{x}\right){( 1- \text{x})}^{2}}
\left(2 + $$\text{x}$\right) \to2($\text{x}$ is small)
\left(1 - \text{x}\right) \to1
{K}_{P} = \dfrac{4x^{3} \text{p}}{8} = \dfrac{x^{3} \text{p}}{2}
$\text{x}$ =\dfrac{2 $\text{kp}$}{$p}3