I suspect the answer to this question is no in general.
Suppose $Z$ has two components $Z_1 \cup Z_2$ with different multiplicities (as a formal linear combination of knots) and $Z'$ has one component. Then $P$ must have two points $P=\{p_1,p_2\}$ (ignoring multiplicity), and $P'$ one $P'=\{p'\}$. Then $\pi_1(M-(Z\cup Z'))$ surjects a free group of rank at least two (the meridians of $Z_i$ go to loops about $p_i$, which must generate a subgroup of $\pi_1(S-(P\cup P'))$ of rank at least two).
The corank of a group is the largest rank of a free group that it surjects. It is known that there are 3-manifold groups with arbitrarily large $b_1$ and corank 1. What is needed to get a counterexample to your question is a manifold with 3 torus boundary components which has fundamental group of corank 1, and satisfying an appropriate homological condition. I feel like a construction of the sort that Shelly Harvey carries out ought to have the right properties.