Let $z_1,z_2,\dots z_n$ are roots. We have $z_1z_2\dots z_n=1$. Therefore $z_k=exp(iy_k), \sum_k y_k=0 (2\pi m), y_k$ are real. It give $a_{n-1}=(-1)^{n-1}(\sum \frac{1}{z_k})=(-1)^n(a_1)^*, a_{n-j}=(-1)^n(a_j)^*,j=1,\dots,n$. Therefore $P(-1)=\frac{(-1)^n}{2} \sum (a_k+(a_k)^*)$ is real.

Cool, but your proof is a little bit obscure and I understood it only from my solution. Actually I did the same just for clearness I proved that: $Im(z_{i_{1}}\cdot z_{i_{2}}\cdot... z_{i_{k}}+z_{j_{1}}\cdot z_{j_{2}}\cdot... z_{j_{n-k}})=0$, where $\{i_{1}, i_{2},..., i_{k}\}+\{j_{1}, j_{2},...,j_{n-k}\}=\{1,2,...,n\}$.

Another formulation of the same idea: let $z_k$ be the roots. Since $z_1z_2\ldots z_n=1$ it follows that $|z_k|=1$ for all $k$ and then $\overline{z_k}=\frac{1}{z_k}$. Now, $P(x)=(x-z_1)\ldots (x-z_n)$ hence $P(-1)=(-1)^n(1+z_1)\ldots (1+z_n).$ Then $\overline{P(-1)}=(-1)^n(1+\overline{z_1})\ldots (1+\overline{z_n})=(-1)^n(1+\frac{1}{z_1})\ldots (1+\frac{1}{z_n})$. We obtain $\overline{P(-1)}=\frac{P(-1)}{z_1\ldots z_n}=P(-1)$. Hence $P(-1)$ is real.