Given is a concentrated solution of a weak electrolyte $$A_xB_y$$ of concentration 'c' and dissociation constant 'K'. The degree of dissociation is given by :

Consider the dissociation equilibrium of the weak electrolyte

$$A_xB_y\rightleftharpoons xA^{y+}+yB^{x-}.$$

Let the initial concentration of the electrolyte be (c) and its degree of dissociation be (\alpha).

At equilibrium,

$$[A_xB_y]=c(1-\alpha),$$

$$[A^{y+}]=xc\alpha,$$

$$[B^{x-}]=yc\alpha.$$

The dissociation constant is

$$K=\frac{[A^{y+}]^x[B^{x-}]^y}{[A_xB_y]}.$$

Substituting the equilibrium concentrations,

$$K=\frac{(xc\alpha)^x(yc\alpha)^y}{c(1-\alpha)}.$$

Since the electrolyte is weak,

$$\alpha\ll1,$$

so that

$$(1-\alpha)\approx1.$$

Therefore,

$$K=\frac{x^xc^x\alpha^x;y^yc^y\alpha^y}{c},$$

$$K=x^xy^yc^{x+y-1}\alpha^{x+y}.$$

Rearranging,

$$\alpha^{x+y}=\frac{K}{x^xy^yc^{,x+y-1}}.$$

Taking the ((x+y)^{\text{th}}) root on both sides,

$$\boxed{\alpha=\left(\frac{K}{x^xy^yc^{,x+y-1}}\right)^{\frac{1}{x+y}}.}$$

Hence, the correct answer is Option B.