The correct answer is $\boxed{\text{A. }1, 1, -1}$.
Let $v$ be an eigenvector of $P$ corresponding to the eigenvalue $\lambda$. Then, $P v = \lambda v$. Multiplying both sides by $P$, we get $P^2 v = P \lambda v = \lambda^2 v$. Continuing in this way, we get $P^3 v = \lambda^3 v$. But since $P^3 = P$, we have $\lambda^3 v = P v = \lambda v$. Therefore, $\lambda^3 = \lambda$.
Since $P$ is a 3 by 3 matrix, its eigenvalues are roots of the characteristic polynomial $p(x) = |xI – P|$. The characteristic polynomial of $P$ is $p(x) = x^3 – x$. The roots of $p(x)$ are $x = 1, 1, -1$. Therefore, the eigenvalues of $P$ are $1, 1, -1$.
Option B is incorrect because it has three distinct eigenvalues. Option C is incorrect because it has two complex eigenvalues. Option D is incorrect because it has one zero eigenvalue and two distinct eigenvalues.