Let $△,\nabla \in \{\wedge ,\vee \}$ be such that $(\mathrm{p}\to \mathrm{q})△(\mathrm{p}\nabla \mathrm{q})$ is a tautology. Then 

Given,

$△,\nabla \in \{\wedge ,\vee \}$ be such that $(\mathrm{p}\to \mathrm{q})△(\mathrm{p}\nabla \mathrm{q})$ is a tautology

Now checking all options by making truth table we get,

Option $\left(\mathrm{A}\right)$ $△=\wedge ,\nabla =\vee$

Option $\left(\mathrm{B}\right)$ $△=\vee ,\nabla =\wedge$

Option $\left(\mathrm{C}\right)$ $△=\vee ,\nabla =\vee$

Hence, it is a tautology.

Option $\left(\mathrm{D}\right)$ $△=\wedge ,\nabla =\wedge$