JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Mathematical Reasoning *

Since $q$ is false, $p\wedge q$ is false whatever $p$ is, and for the biconditional $(p\wedge q)\leftrightarrow r$ to be true $r$ must also be false. So $q$ and $r$ are both false while $p$ is free. Now examine $(p\wedge r)\to (p\vee r)$: with $r$ false, $p\wedge r$ is false, and a conditional with a false antecedent is true regardless of $p$. The statement is true in every allowed case, so it is a tautology.