A graph $G$ is $P_k$-free if and only if $G$ does not have an induced subgraph isomorphic to a path of $k$ vertices. Thus, $P_2$-free graphs are exactly independent sets (or stable sets). $P_4$-free graphs are exactly cographs.
Question. In the worst case, how many $P_3$-free graphs are necessarily needed to partition a $P_4$-free graph of $n$ vertices?
A trivial upperbound is $\frac{n}{2}$. Because graphs of 2 vertices are trivially $P_3$-free, so we can simply partition any graph of $n$ vertices into $\frac{n}{2}$ components with each component having 2 vertices.
A clique-cover number is also an upperbound because clique is also $P_3$-free.