I'll give a counterexample to the original, a possible typo that is true, and then a direct answer to your simplification question.
Cex: Let $G$ be cyclic of order 6, let $P =\{ 2, 3 \}$, and $π = \{ 2 \}$. Then $O^P( G) = 1$ is the identity subgroup, since $G$ itself is a solvable $P$-group. Of course, $O^π(O^P( G ) ) = O^π( 1 ) = 1$ as well. However, $O^π(G )$ is cyclic of order 3, so that the quotient is a $π$-group (a 2-group).
True: On the other hand, I believe $O^P(O^π( G ) ) = O^P(G )$ in general, since solvable $P$-groups is a class closed under extensions. In more detail: $G/O^π(G)$ is a solvable $π$-group, so also a solvable $P$-group. $O^π(G) / O^P(O^π(G) )$ is by definition a solvable $P$-group, and since an extension of a solvable $P$-group by a solvable $P$-group is a solvable $P$-group, one has $G/O^P(O^π(G))$ is a solvable $P$-group. Hence $O^P(G) ≤ O^P(O^π(G))$. Since solvable $P$-groups are closed under normal subgroups, one has $O^P(O^π(G)) ≤ O^P(G)$. Hence $O^P(O^π(G)) = O^P(G)$.
Simplify: It might be that $O^π(O^P(G)) = O^P(G)$. One has $≤$ by definition. $O^P(G)/O^π(O^P(G))$ is a solvable $π$-group, so also a solvable $P$-group, so $G/O^π( O^P(G))$ is a solvable $P$-group, and $O^P(G) ≤ O^π(O^P(G))$.
Note that "true" and "simplify" look similar, but that the hypotheses on $π$ and $P$ are not symmetric.
These should apply to any normal-subgroup and extension closed formations.