Solution:

Since the coefficients of $P$ are real, the complex conjugates of the roots of $P$ are also roots of $P$. Now, if $x$ is a root of unity, then $x^{-1} = \bar{x}$. But the roots of
$$x^{n} P\left(x^{-1}\right) = a_{0} x^{n} + a_{1} x^{n-1} + \cdots + a_{n}$$
are then just the complex conjugates of the roots of $P$, so they are the roots of $P$. Therefore, $P(x)$ and $x^{n} P\left(x^{-1}\right)$ differ by a constant multiple $c$. Since $a_{n} = c a_{0}$ and $a_{0} = c a_{n}$, $c$ is either $1$ or $-1$. But if it were $-1$, then
$$P(1) = a_{n} + a_{n-1} + \cdots + a_{0} = \frac{1}{2}\left(\left(a_{n} + a_{0}\right) + \left(a_{n-1} + a_{1}\right) + \cdots + \left(a_{0} + a_{n}\right)\right) = 0,$$
a contradiction. Therefore $c = 1$, giving the result.