Polynomial with no real roots implies that $\det(P(A))\ge 0$
Since $P$ is real and has no real roots we see that $n$ is even. Since $A$ is real we see that it has an even number of real eigenvalues and the remainder are conjugate pairs.
If $R$ are the indices of the real eigenvalues of $A$ then we see that $\Pi_{k \in R} P(\lambda_k) >0$ (since there are an even number) and if $\lambda_k$ is not real, we see that $P(\lambda_k) P(\overline{\lambda_k}) = P(\lambda_k) \overline{ P(\lambda_k) } = |P(\lambda_k)|^2$. Hence $\det P(A) \ge 0$.