JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Mathematical Reasoning *

Take both connectives to be the implication, giving $(p\wedge q)\to (p\to q)$. If $p\wedge q$ is true, then $p$ and $q$ are both true, so $p\to q$ is true and the whole statement is true; if $p\wedge q$ is false, an implication with a false antecedent is true. The statement holds in all four rows of the truth table, so it is a tautology.
No other pairing works: $(p\wedge q)\wedge (p\to q)$ and $(p\wedge q)\wedge (p\vee q)$ are both false when $p$ is false, and $(p\wedge q)\vee (p\to q)$ is false when $p$ is true and $q$ is false.
So $\odot$ and $\otimes$ are both the implication $\to$.