Explanation

(c)
The weak electrolyte $A_{x}B_{y}$ dissociates as follows
$A_{x}B_{y} \rightleftharpoons xA^{y +} + yB^{x -}\$
C 0 0
$C(1 - \alpha)\ \ \ \ \ xC_{\alpha}\ \ \ \ \ \ \ \ \ \ \ yC_{\alpha}$
Where, $\alpha$= degree of dissociation
$C =$ concentration
$K_{eq} = \frac{\left\lbrack A^{y +} \right\rbrack^{x}\left\lbrack B^{x -} \right\rbrack^{y}}{\lbrack A_{x}B_{y}\rbrack}$
$= \frac{\lbrack xC\alpha\rbrack^{x}\lbrack yC\alpha\rbrack^{y}}{C(1 - \alpha)}$
$= \ \frac{x^{x}.C^{x}\alpha^{x}.y^{y}.C^{y}.a^{y}}{C}$ [$\because 1 - \alpha \approx 1$]
$= x^{x}.y^{y}.\alpha^{x + y}.C^{x + y - 1}$
$a^{x + y} = \frac{K_{eq}}{x^{x}.y^{y}.C^{x + y - 1}}$
$\alpha = \left( \frac{K_{eq}}{x^{x}.y^{y}.C^{x + y - 1}} \right)^{\left( \frac{1}{x + y} \right)}$