In general, it holds that $AB$ and $BA$ have the same non-zero eigenvalues. Thus, $PAP^{-1} = (PA)P^{-1}$ has the same non-zero eigenvalues as $P^{-1}(PA) = (P^{-1}P)A$.
$J := P^{-1}P$ is the orthogonal projection onto the column-space of $P$. If $P$ has full row-rank, then $J$ is the identity matrix. If $P$ has full column-rank, then $P^{-1} = (P^*P)^{-1}P^*$ and it can be shown that $PAP^{-1}$ has the same rank as $P^{-1}PA$.
With the Cauchy interlacing theorem, we can say a bit more. Let $U$ be a matrix with orthonormal columns (i.e. $U^*U = I$) for which $J = UU^*$. Note that $U$ has size $N \times r$, where $r$ is equal to the rank of $P$. We note that $JA = (UU^*)A$ has the same non-zero eigenvalues as the $r \times r$ matrix
$$
U^*AU.
$$
$U^*AU$ is a compression of the matrix $A$, so its eigenvalues interlace the eigenvalues of $A$.
Concretely, we can reach the following two conclusions:
If $P$ has full row-rank $M> N$, then each eigenvalue of $A$ is also an eigenvalue of $PAP^{-1}$ and $0$ is necessarily an eigenvalue of $PAP^{-1}$ (regardless of whether $A$ has $0$ as an eigenvalue). If $A$ has at least one non-negative eigenvalue, then $PAP^{-1}$ and $A$ have the same maximal eigenvalue.
If $P$ has full column-rank $N > M$, then the eigenvalues of $PAP^{-1}$ interlace those of $A$. In particular, if $\mu_1 \leq \cdots \leq \mu_M$ are the eigenvalues of $PAP^{-1}$ and $\lambda_1 \leq \cdots \leq \lambda_N$ are the eigenvalues of $A$, then we have
$$
\lambda_M \leq \mu_M \leq \lambda_N.
$$
