If the percentage dissociation of N₂O₄ is 50, the ratio of Kp and P_eq for N₂O₄ $\rightleftharpoons$2 NO₂  becomes equal to

Total moles at equilibrium = 0.5 +1 = 1.5

Partial pressure of N₂O₄ = ${\mathrm{\chi }}_{{\mathrm{N}}_2{\mathrm{O}}_4}\times {\mathrm{P}}_{\mathrm{eq}}$=$\frac{0.5\times {\mathrm{P}}_{\mathrm{eq}}}{1.5}=\frac{{\mathrm{P}}_{\mathrm{eq}}}{3}$

Partial pressure of NO₂ = ${\mathrm{\chi }}_{{\mathrm{NO}}_2}\times {\mathrm{P}}_{\mathrm{eq}}$ = $\frac{1\times {\mathrm{P}}_{\mathrm{eq}}}{1.5}=\frac{2{\mathrm{P}}_{\mathrm{eq}}}{3}$

 K_p = $\frac{{\left[{\mathrm{P}}_{{\mathrm{NO}}_2}\right]}^2}{\left[{\mathrm{P}}_{{\mathrm{N}}_2{\mathrm{O}}_4}\right]}=\frac{{\displaystyle \frac{4{\mathrm{P}}_{\mathrm{eq}}^2}{9}}}{{\displaystyle \frac{{\mathrm{P}}_{\mathrm{eq}}}{3}}}=\frac{4{\mathrm{P}}_{\mathrm{eq}}}{3}$

$\frac{{\mathrm{K}}_{\mathrm{p}}}{{\mathrm{P}}_{\mathrm{eq}}}=\frac{4}{3}$