smc

We will test each option to see if it can be true or not. Links to diagrams are provided. $$\text{(A)}p-q\text{ can be equal to }\overline{PQ}$$ Let circle $Q$ be inside circle $P$ and tangent to circle $P$, and the point of tangency be $R$. $PR=p$, and $QR=q$, so $PR-QR=PQ=p-q.$ $$\text{(B)}p+q\text{ can be equal to }\overline{PQ}$$ Let circle $Q$ be outside circle $P$ and tangent to circle $P$, and the point of tangency be $R$. $PR=p$, and $QR=q$, so $PR+QR=PQ=p+q.$ $$\text{(C)}p+q\text{ can be less than }\overline{PQ}$$ Let circle $Q$ be outside circle $P$ and not tangent to circle $P$, and the intersection of $\overline{PQ}$ with the circles be $R$ and $S$ respectively. $PR=p$ and $QS=q$, and $PR+QS<PQ$, so $p+q<PQ.$ $$\text{(D)}p-q\text{ can be less than }\overline{PQ}$$ Let circle $Q$ be inside circle $P$ and not tangent to circle $P$, and the intersection of $\overline{PQ}$ with the circles be $R$ and $S$ as shown in the diagram. $PR=p$ and $QS=q$, and $QS<QR$, so $PR-QS<PR-QR$, and $PR-QR=PQ$, so $p-q<PQ.$ Since options A, B, C, and D can be true, the answer must be $\boxed{\text{none of these}}$.