Answer
$\alpha=\sqrt{\frac{K_P}{4+K_P}}$
Work Step by Step
It is to be shown that as long as the extent of the reaction, $\alpha$, for the disassociation reaction $\mathrm{X}_2 \Leftrightarrow 2 \mathrm{X}$ is smaller than one, $\alpha$ is given by $\alpha=\sqrt{\frac{K_P}{4+K_P}}$
The stoichiometric and actual reactions can be written as
Stoichiometric: $\mathrm{X}_2 \Leftrightarrow 2 \mathrm{X}$ (thus $v_{\mathrm{X} 2}=1$ and $v_{\mathrm{X}}=2$ )
Actual:
$$
\mathrm{X}_2 \Leftrightarrow \underbrace{(1-\alpha) \mathrm{X}_2}_{\text {react. }}+\underbrace{2 \alpha X}_{\text {prod. }}
$$ The equilibrium constant $K_p$ is given by
$$
K_p=\frac{N_{\mathrm{X}}^{v_{\mathrm{X}}}}{N_{\mathrm{X} 2}^{v_{\mathrm{X} 2}}}\left(\frac{P}{N_{\text {total }}}\right)^{v_{\mathrm{X}}-v_{\mathrm{X} 2}}=\frac{(2 \alpha)^2}{(1-\alpha)}\left(\frac{1}{\alpha+1}\right)^{2-1}=\frac{4 \alpha^2}{(1-\alpha)(1+\alpha)}
$$ Solving this expression for $\alpha$ gives $$
\alpha=\sqrt{\frac{K_P}{4+K_P}}
$$
