$$A_g \rightleftharpoons B_g + \frac{C}{2}_g$$. The correct relationship between $$K_P$$, $$\alpha$$ and equilibrium pressure $$P$$ is

For the gaseous equilibrium $$A_g \rightleftharpoons B_g + \frac12\,C_g$$ let the initial amount of $$A$$ be 1 mol at an equilibrium (total) pressure $$P$$ and volume $$V$$.

Degree of dissociation $$\alpha$$: fraction of the initial $$A$$ that dissociates.

Number of moles at equilibrium:
$$n_A = 1-\alpha$$ (undissociated $$A$$)
$$n_B = \alpha$$
$$n_C = \dfrac{\alpha}{2}$$

Total moles
$$n_{\text{tot}} = 1-\alpha + \alpha + \dfrac{\alpha}{2} = 1 + \dfrac{\alpha}{2}$$ $$-(1)$$

Partial pressures (using $$P_i = \dfrac{n_i}{n_{\text{tot}}}\,P$$):
$$P_A = \dfrac{1-\alpha}{1+\alpha/2}\;P$$
$$P_B = \dfrac{\alpha}{1+\alpha/2}\;P$$
$$P_C = \dfrac{\alpha/2}{1+\alpha/2}\;P = \dfrac{\alpha}{2(1+\alpha/2)}\;P$$

Expression for $$K_P$$ (pressure equilibrium constant):
$$K_P = \dfrac{P_B\,P_C^{1/2}}{P_A}$$ $$-(2)$$

Substituting the partial pressures from above into $$(2)$$:

$$\begin{aligned} K_P &= \dfrac{\displaystyle \frac{\alpha P}{1+\alpha/2}\; \left[\frac{\alpha P}{2(1+\alpha/2)}\right]^{1/2}} {\displaystyle \frac{(1-\alpha)P}{1+\alpha/2}} \\[6pt] &= \dfrac{\alpha P}{1+\alpha/2}\; \dfrac{\sqrt{\alpha P}}{\sqrt{2(1+\alpha/2)}}\; \dfrac{1+\alpha/2}{(1-\alpha)P} \\[6pt] &= \dfrac{\alpha^{3/2}P^{3/2}} {(1+\alpha/2)^{3/2}\,\sqrt2}\; \dfrac{1}{(1-\alpha)P}\; (1+\alpha/2) \\[6pt] &= \dfrac{\alpha^{3/2}P^{1/2}} {\sqrt2\,(1-\alpha)\,\sqrt{1+\alpha/2}} \\[6pt] &= \dfrac{\alpha^{3/2}P^{1/2}} {(1-\alpha)\,\sqrt{(2+\alpha)/2}} \\[6pt] &= \dfrac{\alpha^{3/2}P^{1/2}} {(1-\alpha)\,(\sqrt{2+\alpha}/\sqrt2)} \\[6pt] &= \dfrac{\alpha^{3/2}P^{1/2}} {(\sqrt{2+\alpha})(1-\alpha)}. \end{aligned}$$

Therefore

$$K_P = \dfrac{\alpha^{3/2}P^{1/2}} {(2+\alpha)^{1/2}(1-\alpha)}$$

Comparing with the given options, this matches Option B.

Answer : Option B