Expressing a root of an irreducible polynomials using the other roots

The following is too long for a comment, but I felt it adds some clarity to the question, even though it does not provide an answer:

Let $G = \text{Gal}_{\mathbb{Q}}(K_P)$, let $F = \mathbb{Q}(r_1)$ and $H = \text{Gal}_F(K_P)$. Then, by this answer, your question is equivalent to asking

Does there exist a solvable subgroup $L < G$ such that $\langle LH\rangle = G$.

What we know:

1. Consider the injective homomorphism $$ G\hookrightarrow S_n $$ via the (faithful, transitive) action of $G$ on the roots $X:= \{r_1,r_2,\ldots, r_n\}$, then it follows that $$ H = G\cap S_{n-1} $$
2. $[G:H] = [\mathbb{Q}(r_1):\mathbb{Q}] = \deg(P(x)) = n$, since $P$ is separable.
3. If $G$ is solvable, there is nothing to prove, so we may assume $n\geq 5$.
4. If $G = S_n$, then $L = \langle (12)\rangle$ works.
5. If $G = A_n$, then (I suspect), $L = \langle (12)(34)\rangle$ works.
6. If $n=5$, the only possibilities for $G$ are $S_5, A_5$ and 3 solvable groups ($\mathbb{Z}_5, D_5$, and $F_{20}$), so the result you want is true in this case.
7. These notes of Keith Conrad might be useful, particularly Theorem 2.2