this isn't quite what was desired but the problem is still trivial

$(PQ)^{-1}=Q^{-1}P^{-1}=(QR)^{-1}=R^{-1}Q^{-1}$

Multiply both sides on the left by $Q$ to get

$P^{-1}=Q R^{-1}Q^{-1}$

for the rest of it just multiply again by $P^{-1}$ and do some more matrix manipulations, it's not hard.

Well call $QR$, by $A$. Using the hint:

$(AQ)^{-1} = (Q^{-1})^{-1}A^{-1} = QA^{-1}$.

Using the hint again:

$A^{-1} = (QR)^{-1} = R^{-1}Q^{-1}$, so: $P^{-1} = QR^{-1}Q^{-1}$.

If one knows that $GL_n(V)$ is a group, one can simply verify that:

$P(QR^{-1}Q^{-1}) = (QRQ^{-1})(QR^{-1}Q^{-1}) = (QR)(Q^{-1}Q)(R^{-1}Q^{-1}) = (QR)(I)(R^{-1}Q^{-1}) = (QR)(R^{-1}Q^{-1}) = Q(RR^{-1})Q^{-1} = QIQ^{-1} = QQ^{-1} = I$.

and conclude $P^{-1} = QRQ^{-1}$ from the uniqueness of inverses.

In fact, it can be shown that $QAQ^{-1}QBQ^{-1} = Q(AB)Q^{-1}$, for any square matrices $A,B$ of the proper size and any invertible $Q$ of matching size. This, in turn, leads to $(QAQ^{-1})^k = QA^kQ^{-1}$.