It is correct. You could simplify the check by using some “propositional identities”: for instance, you can prove that
$$
A\land(B\lor C)\equiv (A\land B)\lor(A\land C)
$$
so in your case you can substitute $\lnot Q\land(\lnot P\lor Q)$ with
$$
(\lnot Q\land\lnot P)\lor(\lnot Q\land Q)
$$
Since $\lnot Q\land Q$ is always false, you can substitute the whole with $\lnot Q\land \lnot P$. Thus the first block becomes (equivalent to)
$$
P\land(\lnot Q\land\lnot P)
$$
Oh, well, this is false for every $P$ and $Q$, because of
$$
X\land Y\equiv Y\land X
\qquad\text{and}\qquad
A\land(B\land C)\equiv (A\land B)\land C
$$
so
$$
P\land(\lnot Q\land\lnot P)\equiv
P\land(\lnot P\land\lnot Q)\equiv
(P\land\lnot P)\land\lnot Q
$$
It is correct. You could simplify the check by using some “propositional identities”: for instance, you can prove that $$ A\land(B\lor C)\equiv (A\land B)\lor(A\land C) $$ so in your case you can substitute $\lnot Q\land(\lnot P\lor Q)$ with $$ (\lnot Q\land\lnot P)\lor(\lnot Q\land Q) $$ Since $\lnot Q\land Q$ is always false, you can substitute the whole with $\lnot Q\land \lnot P$. Thus the first block becomes (equivalent to) $$ P\land(\lnot Q\land\lnot P) $$ Oh, well, this is false for every $P$ and $Q$, because of $$ X\land Y\equiv Y\land X \qquad\text{and}\qquad A\land(B\land C)\equiv (A\land B)\land C $$ so $$ P\land(\lnot Q\land\lnot P)\equiv P\land(\lnot P\land\lnot Q)\equiv (P\land\lnot P)\land\lnot Q $$