At temperature $$T$$, compound $$AB_{2(g)}$$ dissociates as $$AB_{2} \rightleftharpoons AB_{(g)}+\frac{1}{2}B_{2(g)}$$ having degree of dissociation $$x$$ (small compared to unity). The correct expression for $$x$$ in terms of $$K_{p}$$ and p is

For $$AB_2 \rightleftharpoons AB + \frac{1}{2}B_2$$ with small degree of dissociation $$x$$:

Starting with 1 mole at pressure $$p$$, at equilibrium: $$AB_2$$: $$1-x$$, $$AB$$: $$x$$, $$B_2$$: $$x/2$$.

Total moles $$\approx 1$$ (since $$x \ll 1$$). Partial pressures: $$p_{AB} \approx xp$$, $$p_{B_2} \approx xp/2$$, $$p_{AB_2} \approx p$$.

$$K_p = \frac{p_{AB} \cdot p_{B_2}^{1/2}}{p_{AB_2}} \approx \frac{xp \cdot (xp/2)^{1/2}}{p} = \frac{x^{3/2} \sqrt{p}}{\sqrt{2}}$$

Squaring: $$K_p^2 = \frac{x^3 p}{2}$$, so $$x^3 = \frac{2K_p^2}{p}$$.

$$x = \left(\frac{2K_p^2}{p}\right)^{1/3} = \sqrt[3]{\frac{2K_p^2}{p}}$$

The correct answer is Option 3.