jmc

Let's start with a picture: We can carve $ABCD$ into four (non-regular) tetrahedra that share $P$ as a vertex and have respective bases $ABC$, $ABD$, $ACD$, and $BCD$ (the faces of $ABCD$). For example, this diagram shows one of these four tetrahedra, namely $BCDP$: The four tetrahedra formed in this way are congruent, so each contains one-quarter the volume of $ABCD$.

The height of tetrahedron $BCDP$ is $PQ$, so the volume of $BCDP$ is $$\frac{1}{3}\cdot (\text{area of }△BCD)\cdot PQ.$$The volume of the original tetrahedron, $ABCD$, is $$\frac{1}{3}\cdot (\text{area of }△BCD)\cdot AQ.$$Thus $PQ/AQ$ is equal to the ratio of the volume of $BCDP$ to the volume of $ABCD$, which we already know to be $\boxed{\frac{1}{4}}$.