The equilibrium constant \( K_p \) for the reaction is given by:
\[
K_p = \frac{p_{X_2} \cdot p_{Y_2}^{1/2}}{p_{X_2Y}}.
\]
Using the degree of dissociation \( x \) for the reactant \( X_2Y \), the partial pressures at equilibrium are:
\[
p_{X_2Y} = p_0 - x \cdot p_0, \quad p_{X_2} = x \cdot p_0, \quad p_{Y_2} = \frac{x \cdot p_0}{2}.
\]
Substituting these into the expression for \( K_p \) and simplifying for small \( x \), we get:
\[
x = \sqrt{\frac{2K_p}{p}}.
\]
Thus, the correct answer is \( x = \sqrt{\frac{2K_p}{p}} \).