For a reaction at equilibrium
$$A(g) \rightleftharpoons B(g) + \frac{1}{2}C(g)$$
the relation between dissociation constant ($$K$$), degree of dissociation ($$\alpha$$) and equilibrium pressure ($$p$$) is given by :

For the equilibrium reaction: $$A(g) \rightleftharpoons B(g) + \frac{1}{2}C(g)$$ we start with 1 mole of A. At equilibrium, the moles of A, B, and C are $$1-\alpha$$, $$\alpha$$, and $$\frac{\alpha}{2}$$ respectively, giving a total of $$n_{\text{total}} = (1 - \alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2}$$.

The mole fractions yield the partial pressures (total pressure $$p$$) as $$p_A = \frac{1 - \alpha}{1 + \frac{\alpha}{2}} \cdot p$$, $$p_B = \frac{\alpha}{1 + \frac{\alpha}{2}} \cdot p$$, and $$p_C = \frac{\frac{\alpha}{2}}{1 + \frac{\alpha}{2}} \cdot p = \frac{\alpha}{2\left(1 + \frac{\alpha}{2}\right)} \cdot p$$.

$$K = \frac{p_B \cdot p_C^{1/2}}{p_A}$$

$$K = \frac{\frac{\alpha p}{1 + \frac{\alpha}{2}} \cdot \left(\frac{\alpha p}{2(1 + \frac{\alpha}{2})}\right)^{1/2}}{\frac{(1 - \alpha)p}{1 + \frac{\alpha}{2}}}$$

$$K = \frac{\alpha p}{\left(1 + \frac{\alpha}{2}\right)} \cdot \frac{\alpha^{1/2} p^{1/2}}{2^{1/2}\left(1 + \frac{\alpha}{2}\right)^{1/2}} \cdot \frac{\left(1 + \frac{\alpha}{2}\right)}{(1 - \alpha)p} = \frac{\alpha^{3/2} \cdot p^{1/2}}{2^{1/2} \cdot \left(1 + \frac{\alpha}{2}\right)^{1/2} \cdot (1 - \alpha)}$$

Noting that $$2^{1/2} \cdot \left(1 + \frac{\alpha}{2}\right)^{1/2} = \left(2\left(1 + \frac{\alpha}{2}\right)\right)^{1/2} = (2 + \alpha)^{1/2}$$, we have

$$K = \frac{\alpha^{3/2} \cdot p^{1/2}}{(2 + \alpha)^{1/2} \cdot (1 - \alpha)}$$

The correct answer is Option A: $$K = \frac{\alpha^{3/2} p^{1/2}}{(2+\alpha)^{1/2}(1-\alpha)}$$.