Consider the following data.

Electrolyte	$$\Lambda^\circ_m$$ (S cm$$^2$$ mol$$^{-1}$$)
BaCl$$_2$$	$$x_1$$
H$$_2$$SO$$_4$$	$$x_2$$
HCl	$$x_3$$

BaSO$$_4$$ is sparingly soluble in water. If the conductivity of the saturated BaSO$$_4$$ solution is $$x$$ S cm$$^{-1}$$ then the solubility product of BaSO$$_4$$ can be given as
(Here $$\Lambda_m = \Lambda^\circ_m$$)

According to Kohlrausch's Law, the limiting molar conductivity of an electrolyte is equal to the sum of the limiting ionic conductivities of its constituent ions.

Given,

$$\Lambda_m^0(BaCl_2)=x_1=\lambda_{Ba^{2+}}^0+2\lambda_{Cl^-}^0,$$

$$\Lambda_m^0(H_2SO_4)=x_2=2\lambda_{H^+}^0+\lambda_{SO_4^{2-}}^0,$$

$$\Lambda_m^0(HCl)=x_3=\lambda_{H^+}^0+\lambda_{Cl^-}^0.$$

Hence,

$$\Lambda_m^0(BaSO_4)=\lambda_{Ba^{2+}}^0+\lambda_{SO_4^{2-}}^0,$$

$$\Lambda_m^0(BaSO_4)=\Lambda_m^0(BaCl_2)+\Lambda_m^0(H_2SO_4)-2\Lambda_m^0(HCl),$$

$$\Lambda_m^0(BaSO_4)=x_1+x_2-2x_3.$$

For a sparingly soluble salt,

$$\Lambda_m^0=\frac{1000,\kappa}{S},$$

where (\kappa=x\ \text{S cm}^{-1}) and (S) is the solubility in (\text{mol L}^{-1}).

Therefore,

$$S=\frac{1000x}{\Lambda_m^0(BaSO_4)},$$

$$S=\frac{1000x}{x_1+x_2-2x_3}.$$

Since

$$BaSO_4(s)\rightleftharpoons Ba^{2+}(aq)+SO_4^{2-}(aq),$$

the solubility product is

$$K_{sp}=[Ba^{2+}][SO_4^{2-}]=S^2.$$

Substituting the value of (S),

$$K_{sp}=\left(\frac{1000x}{x_1+x_2-2x_3}\right)^2,$$

$$K_{sp}=\frac{10^6x^2}{(x_1+x_2-2x_3)^2}.$$

Hence, the correct expression for the solubility product is

$$K_{sp}=\frac{10^6x^2}{(x_1+x_2-2x_3)^2}.$$