Answer: C
We know that $a \rightarrow b = \lnot a \lor b$
So, $((R∨Q)∧(P∨∼Q))= ((\lnot R \rightarrow Q) \land (Q \rightarrow P))$, if this is true then
both $(\lnot R \rightarrow Q)$ and $(Q \rightarrow P)$ must be true and from this we can conclude that $(\lnot R \rightarrow P)$
We can write $(\lnot R \rightarrow P) = (R \lor \lnot P)$
Now we know that when $A \land B \rightarrow C$ so if $A \land B$ is true so $C$ must be true and when $A \land B$ evaulates to false we don’t care about truth value of $C$ because expression will ultimately evaluates to false.
So, $((R∨Q)∧(P∨∼Q))$ and $((R∨Q)∧(P∨∼Q) \land (R \lor \lnot P))$ will have same truth table hence both of these expression is equivalent.