The square map on $SO(n)$

It can't be a covering map for $n>2$, since the fundamental group of $SO(n)$ for $n>2$ is finite, hence it can't cover itself with more than one sheet. But note that $P(I)=I$ and $P(A)=I$, where $$A=\begin{pmatrix} -1 & 0 &0 & \cdots & 0 \\ 0 &-1 & 0 &\cdots &0 \\ 0 &0 & 1 &\cdots & 0 \\ 0 &0 &0 &\cdots &1 \end{pmatrix}.$$

For $n=2$, $SO(n)$ is $S^1$, so the map is a covering map.

For $n=1$, it is the identity, so the map is a covering map.

There are several obstructions to this map being a covering map for $n \ge 3$. A simple one is that its fibers are not discrete: for example, $P^{-1}(I)$ contains all $180^{\circ}$ rotations about every axis.

Alternatively, you can use the fact that a map between closed manifolds is a covering map iff it's a local diffeomorphism, and then compute the differential of $P$ to see where it fails to be a local diffeomorphism. The tangent space at an element $X \in SO(n)$ can be identified with the set of matrices of the form $X(I + \epsilon Y)$ where $Y \in \mathfrak{so}(n)$ is a skew-symmetric matrix and $\epsilon^2 = 0$. Squaring such a matrix gives

$$\left( X + \epsilon XY \right)^2 = X^2 + \epsilon (X^2 Y + XYX) = X^2 \left( I + \epsilon (Y + X^{-1} Y X) \right).$$

So the differential $P_X$ can be identified with the linear map $Y \mapsto Y + X^{-1} Y X$. This is an isomorphism iff it has nontrivial kernel, and its kernel consists of matrices $Y$ satisfying $Y + X^{-1} Y X = 0$.

The matrices $Y$ such that some $X$ exists with this property are matrices conjugate to $-Y$ by an element of $SO(n)$. Every skew-symmetric matrix $Y$ is conjugate to $-Y$, even by an element of $O(n)$, as follows: $Y$ is conjugate via an element of $O(n)$ to a block diagonal matrix whose blocks have the form

$$\left[ \begin{array}{cc} 0 & \theta \\ - \theta & 0 \end{array} \right]$$

for some $\theta$, and any such block is conjugate to its negation via a reflection. The extra condition that we need $Y$ to be conjugate to $-Y$ via an element of $SO(n)$ means we need to do an even number of reflections, which is always possible when $n \ge 4$. When $n = 3$ one of the eigenvalues of $Y$ is zero and so we can just stick a $-1$ into $X$ as necessary.

Hence for any nonzero value of $Y$ we can find $X$ with the above property, and so there are many points at which the differential fails to be surjective and hence when $P$ fails to be a local diffeomorphism.