Let $\Delta ,\nabla \in \left\{\wedge ,\vee \right\}$ be such that $p\nabla q\to \left(\left(p\Delta q\right)\nabla r\right)$ is a tautology. Then $\left(p\nabla q\right)∆r$ is logically equivalent to

Case-I

If $\Delta \equiv \nabla \equiv \wedge$

$\left(p\wedge q\right)\to \left(\left(p\wedge q\right)\wedge r\right)$

then it is false if $r$ is false.

So not a tautology

Case-II

If $\Delta \equiv \nabla \equiv \vee$

$\left(p\vee q\right)\to \left(\left(p\vee q\right)\vee r\right)\equiv$ tautology

then $\left(p\vee q\right)\vee r\equiv \left(p\Delta r\right)\vee q$

Case-III

If $\Delta =\vee ,\nabla =\wedge$

then $\left(p\wedge q\right)\to \left\{\left(p\vee q\right)\wedge r\right\}$ is not a tautology

when $p\to T,q\to T,r\to F$

Case-IV

If $\Delta =\wedge ,\nabla =\vee$

$\left(p\wedge q\right)\to \left\{\left(p\wedge q\right)\vee r\right\}$ is not a tautology