The degree of dissociation $\left(\alpha \right)$ of a weak electrolyte $A_x{B}_y$ is related to Van't Hoff factor (i) by the expression 

$$\begin{gathered} A_x+B_{y }\leftrightarrow x{A}^{y+}+y{B}^{x-} \\ 1-\alpha \alpha x \alpha y \end{gathered}$$
Total amount of species = $1+(x+y-1)\alpha$
Vant Hoff factor = $i = \frac{1+(x+y-1)\alpha }{1}$

$\alpha =\frac{i-1}{(x+y-1)}$

Therefore, Option A is the correct answer