The dissociation of a weak electrolyte ($\mathrm{HA}$) in solution can be represented as:

$ \mathrm{HA}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{A}^{-}(\mathrm{aq}) $

The degree of dissociation ($\alpha$) for a weak electrolyte is given by:

$ \alpha = \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} $

Where $\Lambda_{\mathrm{m}}$ is the molar conductivity at a given concentration $C$, and $\Lambda_{\mathrm{m}}^{\circ}$ is the molar conductivity at infinite dilution.

The dissociation constant ($K_{\mathrm{a}}$) is described by:

$ K_{\mathrm{a}} = \frac{\alpha^2 C}{1 - \alpha} $

Rewriting this equation in terms of $\Lambda_{\mathrm{m}}$ and $\Lambda_{\mathrm{m}}^{\circ}$:

$ K_{\mathrm{a}} = \frac{\left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right)^2 C}{1 - \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}} $

By multiplying out and simplifying the above expression:

$ \left( \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} \right)^2 C + K_{\mathrm{a}} \left( \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} \right) = K_{\mathrm{a}} $

Substitute $\alpha = \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}$:

$ \left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right)^2 C + K_{\mathrm{a}} \left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right) - K_{\mathrm{a}} = 0 $

Rewriting by multiplying through by $\left( \Lambda_{\mathrm{m}}^{\circ} \right)^2$:

$ \Lambda_{\mathrm{m}}^2 C + K_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ} - K_{\mathrm{a}} \left( \Lambda_{\mathrm{m}}^{\circ} \right)^2 = 0 $

Thus, the correct equation that describes the change in molar conductivity with respect to concentration for a weak electrolyte is given by:

Option D:

$ \Lambda_{\mathrm{m}}^2 C - K_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ 2} + K_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ} = 0 $