Let $\Sigma, \Sigma_{1}, \Sigma_{2}$ be respectively the circumcircles of $ABCD, \triangle PBD, \triangle PAC$. Let also $C_{1}, C_{2}$ be the centres of $\Sigma_{1}, \Sigma_{2}$ respectively. Consider the radical axes of each two of them. Since the radical axis of $\Sigma, \Sigma_{1}$ is $BD$ and that of $\Sigma, \Sigma_{2}$ is $AC$, the radical axis of $\Sigma_{1}, \Sigma_{2}$ must be concurrent with $AC$ and $BD$, i.e. it passes through $E$. Therefore, the radical axis of $\Sigma_{1}, \Sigma_{2}$ is $PE$. To prove $O$, $P$, $E$ are collinear, it suffices to prove $O$ has the same power with respect to $\Sigma_{1}$ and $\Sigma_{2}$. Without loss of generality we may let $\angle A$ and $\angle B$ to be not smaller than a right angle. Then point $P$ lies in $\triangle CDE$ and it is outside $\Sigma_{1}$ and $\Sigma_{2}$. By some angle chasing we can prove that $\angle ODC_{1}=90^\circ$, and so the power of $O$ w.r.t $\Sigma_{1}$ is equal to $OC_{1}^{2}-CC_{1}^{2}=OC^{2}$. Similarly, the power of $O$ w.r.t. $\Sigma_{2}$ is also the square of the circumradius of $ABCD$. The result follows.

Consider inversion with center $P$ and any power. Then see that our hypothesis implies that $A^{*}B^{*}C^{*}D^{*}$ ($X^{*}$ denotes point $X$ after inversion) is a rectangle. Let $O'$ be center of circumcircle of $A^{*}B^{*}C^{*}D^{*}$. Point $E^{*}$ is the intersection of circumcircles of $PBD$ and $PCA$ and hence lies on line $PO'$ (as $O'$ is the intersection of $A^{*}C^{*}$ and $B^{*}D^{*}$). But well-known fact says that $P,O',O$ are collinear and hence $P,O',O^{*}$ are collinear, so $P,E^{*},O^{*}$ are collinear and hence $P,E,O$ are collinear. QED

Let $A', B', C', D'$ be the intersection point of the circle and $AP, BP, CP, DP$ respectively. By angle tracing, $A'B'C'D'$ is a rectangle. Therefore $O$ is the intersection point of $A'C'$ and $B'D'$. Use central projection to map $P$ to the centre of the circle. Then by symmetry, $P, E, O$ are collinear.

Diagram: hope that helps a little (It helped me...)

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Another solution: Let $AP$ cut the circle at $F$, $BP$ cut the circle at $H$. $\angle PAB + \angle PCB = \angle PCF = 90^{\circ}$,so $PC$ must cut $OF$ at a point $G$ on the circle. Similarly, $DP$ must cut $OH$ at $K$ on the circle. Let $GB$ cut $AK$ at $X$. By Pascal theorem in $ACGBDK: E = AC \cap BD, P = CG \cap DK, X = AK \cap GB$ are collinear. By Pascal theorem in $GFAKHB: O = GF \cap KH, P = FA \cap HB, X = GB \cap AK$ are collinear. So $E, P, X, O$ are collinear

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If $AP,BP,CP,DP$ cut to $\odot (ABCD)$ at $A',B',C',D'$ respectively. Consider $\mathcal{G}$ the composition of involutions with poles $P,E,P$ that fixed the conic $\odot (ABCD)$ is well-known that is a involution with pole on $EP$. Since $\mathcal{G}(A')=C'$ and $\mathcal{G}(B')=D' \Rightarrow A'C'$ and $B'D'$ are secants at the pole of $\mathcal{G}$. Using the condition $\angle PAB+\angle PCB=\angle PBC+\angle PDC=90^\circ$ we get $A'C'$ and $B'D'$, and are secants at $O$ so $E,P,O$ are collinear.

This was in Korean Postal Coaching.. Let $O_1$ be the pole of $AC$ with respect to $O$. Draw a circle - with $O_1$ as its center and $O_1A=O_1C$ as its radius. Then we have $2\angle APC = 2(90+\angle ADC)=180+\angle AOC = 360-\angle AO_1C$, so $P$ lies on the circle $O_1$. Similarly, $P$ lies on the circle $O_2$. I claim that $O, P, E$ lie on the radical axis of $O_1, O_2$. First, $O$ lies on the radical axis since $P_{O_1}(O)=OA^2=OB^2=P_{O_2}(O)$. Also, $E$ lies on the radical axis since $AE \cdot EC = BE \cdot ED$. Therefore, we are done. $\blacksquare$