Here's a nice solution with inversion. Consider the inversion $I$ centered at $P$ with random radius, and denote , in general, $I(W)=W'$. We will prove that $P$ lies at the intersection of the interior bisectrix of $\angle A$ ,$\angle B$ ,$\angle C$ and $\angle D$. Start by noticing that $\angle APB + \angle CPD=\angle APD + \angle BPC=180^{\circ} $ $(1)$ and that the original conditions become: \begin{align*} \angle PD'A'+ \angle PD'C'=\angle PB'A' + \angle PB'C' \implies \angle C'D'A'=\angle C'B'A' ;\\ \angle PC'D' + \angle PC'B'= \angle PA'D' + \angle PA'B' \implies \angle B'C'D'=\angle B'A'D'. \end{align*}It's easy to see that $A'B'C'D'$ is a parallelogram that still satisfies condition $(1)$, as inversion doesn't affect it. We'll present the following well-known lemma that ultimately kills the problem: $\textcolor{red}{\textbf{Lemma.}}$ Let $XYZT$ be a parallelogram and $O$ be a point in its interior such that $\angle XOY + \angle TOZ=\angle YOZ+ \angle XOT=180^{\circ}$. Than,the following equalities occur: $\angle XTO= \angle XYO$, $\angle ZTO=\angle ZYO$, $\angle YXO= \angle YZO$ and $\angle TXO= \angle TZO$. $\textcolor{blue}{\textbf{Proof.}}$ Let $S$ be a point on the parallel through $O$ to $XT$ and $YZ$ such that the segment $OS$ intersects $XY$ and $OS=XT=YZ$. It suffices to prove the first equality, as the other three result from switching the position of segment $OS$. $XTOS$ and $YZOS$ are parallelograms $\implies \angle SXO+ \angle SYO=\angle XOT +\angle YOZ=180^{\circ}$, so $XOYS$ is cyclic. By simple angle-chasing, one can see that $\angle XYO=\angle XSO=\angle XTO$. $ \square $ The lemma implies that $\angle PD'A'=\angle PB'A'$, $ \angle PA'B'=\angle PC'B' $, $\angle PB'C'= \angle PD'C' $, and $ \angle PC'D'= \angle PA'D'$ $ \overset {I^{-1}} {\implies} \angle PAD=\angle PAB$, $\angle PBA=\angle PBC$, $\angle PCB=\angle PCD$, $\angle PDC=\angle PDA$ $\implies AP$, $BP$, $CP$, $DP$ are the interior bisectrix of $\angle A$, $\angle B$, $\angle C$, and $\angle D$.