In a reaction $A{B}_{\left(g\right)}\underset{ }{\rightleftharpoons }{A}_{\left(g\right)}+B_{\left(g\right)}$
The value of K p is numerically equal to 1/3^rd of total equilibrium pressure. What is the percentage dissociation of AB?

$$\begin{gathered} A\underset{\left(g\right)}{B}\underset{\left(g\right)}{A}+\underset{\left(g\right)}{B} \\ A+t=0 p 0 0 \end{gathered}$$
At equilibrium $p-p^1{p}^1{p}^1$
$\begin{array}{l}\therefore \text{ Total }\mathrm{p}=p-p^1+p^1+p^1\\ =p+p^1\\ K_p=\frac{1}{3}\left(p+p^1\right)\to 1\\ K_p=\frac{P_A{P}_B}{PAB}\\ \frac{1}{3}\left(P+P^1\right)=\frac{P_1^2}{P-P^1}\\ \frac{P^2}{4}=P_1^2\Rightarrow \frac{P}{2}=P^1\end{array}$
50%  is dissociated