In fact the symmetric part of $P$, i.e., $P_S= \tfrac12 (P + P^t)$ can alway be diagonalized by a transformation matrix $A$. In fact it is possible to obtain $ A P_S A^t = \begin{pmatrix} 1 \\ & \ddots\\ & & -1\\ & & &\ddots \\ & & & &0\\ & & & & &\ddots \end{pmatrix} =\Sigma;$ with $\Sigma$ the signature of the symmetric part of $P$ having $n_+$-times $1$ and $n_-$-times $-1$ and $n-n_+-n_-$-times 0 on the diagonal.
On the other hand, the antisymmetric part $P_A = \tfrac12 (P- P^t)$ remains antisymmetric under the congruence relation and thus cannot be diagonalized. We thus have $A P A^t = \Sigma + W$ with $W=-W^t$.
The question whether $A P A^t$ can be diagonalized can now be restated: does a $U\in GL(n, \mathbb{C})$ exist with $U\Sigma U^{-1} = \Sigma$ and $U W U^{-1}$ diagonal which is equivalent to the question if the eigenvectors of generalized eigenvalue problem $ \det(\Sigma- \lambda W) =0$ span the whole $\mathbb{C}^n$.
For two cases the answer is obviously yes. If $n_+ =n$ or $n_-=n$, we have the spectral theorem of skew-symmetric matrices which tells us that in fact $APA^t$ can be diagonalized. Thus if $P_S$ is positive or negative definite then the answer is that there exist a congruent matrix to $P$ which can be diagonalized. For $n_+ = n_- =0$ ($P$ is an antisymmetric matrix) then $\Sigma$ remains obviously invariant and the answer is again yes.
For the other cases we should know whether there exist an $U\in SU(n_+,n_-)$ which diagonalizes a skew-symmetric matrix...