Let $G$ be any finite group. Since $Z(G)$ is an abelian group, it can in particular be decomposed as a product of cyclic groups. So, in particular, we can define $Z_{p'}(G) = \{ g \in Z(G) \; : \; p \text{ does not divide order}(g)\}$ and have that it is well-defined. Essentially, it is a product of all those cyclic groups that appear in the decomposition of $Z(G)$ and have order prime to $p$. Also, we denote by $O_p(G)$ the maximal normal $p$-subgroup of $G$.
I have been trying to prove this statement (if it is true). I believe that I have found a proof, however I want a second opinion. $$Z(G)O_p(G) = Z_{p'}(G)O_p(G)$$
One inclusion is, of course, obvious. About the other direction, if we consider $Z_p(G) = Z(G)/Z_{p'}(G)$, we can take $Z_p(G)O_p(G)$, it is a subgroup of $G$ (because the two groups, obviously, commute) and it is also a normal subgroup of $G$. Since it also contains $O_p(G)$, it follows from the definition of $O_p(G)$ that $Z_p(G)O_p(G) = O_p(G)$, which proves the thesis (that is, you can only consider the $p'$ part of $Z(G)$ and get the same group).