Need to prove $(p \land q) \land (\lnot p \lor r) \rightarrow (q \lor r)$ is a tautology.

$p \land q \land (\lnot p \lor r)$ implies $q$ $\tiny\text{… by simplification}$

$q$ implies $q \lor r$ $\tiny\text{… by addition}$

The rest is obvious.

First note that $$F \lor ( (p \land q) \land r)=p \land q \land r$$ Then recall that How to prove that $P \rightarrow Q$ is equivalent with $\neg P \lor Q $?

Hence, from your work, $$(p \land q)\land ( \lnot p \lor r) \rightarrow (q \lor r)$$ is equivalent to $$\lnot(p \land q \land r)\lor (q \lor r)$$ that is, by using De Morgan's laws, $$(\lnot p \lor \lnot q \lor \lnot r) \lor (q \lor r)=\lnot p \lor (\lnot q \lor q) \lor (\lnot r\lor r)=T.$$