Let $\Omega$ be the circumcircle of the quadrilateral $A B C D$ and let $r$ be its radius.
First, notice that the points $X, Y, Z, W$ are concyclic. Indeed, using oriented (modulo $180^{\circ}$ ) angles,

$$\angle(X W, X Y)+\angle(Z Y, Z W)=\angle(X A, X B)+\angle(Z C, Z D)=-\frac{\angle A+\angle B}{2}-\frac{\angle C+\angle D}{2}=0 .$$

Let $\omega$ be the circle passing through these four points. Our goal is to prove that $O \in \omega$ occurs if and only if $P \in \omega$.

Next, we rule out the case when $A B C D$ is a trapezium. Suppose that if $A B C D$ is an isosceles trapezium; without loss of generality say $A B \| C D$. By symmetry, points $X, Z, O$ and $P$ lie on the symmetry axis (that is, the common perpendicular bisector of $A B$ and $C D$ ), therefore they are collinear. By the conditions of the problem, these four points are distinct and $X$ and $Z$ are on $\omega$, so neither $O$, nor $P$ can lie on $\omega$; therefore the problem statement is obvious. From now on we assume that $A B \nVdash C D$ and $B C \nVdash A D$.
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Let $A B$ and $C D$ meet at $Q$, and let $B C$ and $A D$ meet at $R$. Without loss of generality, suppose that $B$ lies between $A$ and $Q$, and $D$ lies between $A$ and $R$. Now we show that line $Q R$ is the radical axis between circles $\Omega$ and $\omega$.

Point $W$ is the intersection point of the bisectors of $\angle A$ and $\angle D$, so in triangle $A D Q$, point $W$ is the incentre. Similarly, $B Y$ and $C Y$ are the external bisectors of $\angle Q B C$ and $\angle B C Q$, so in triangle $B C Q$, point $Y$ is the excentre opposite to $Q$. Hence, both $Y$ and $W$ lie on the bisector of $\angle D Q A$.

By $\angle D Q A=180^{\circ}-\angle D-\angle A=\angle B-\angle A$ we get

$$\angle B Y Q=\angle Y B A-\angle Y Q B=\frac{\angle B}{2}-\frac{\angle D Q A}{2}=\frac{\angle B}{2}-\frac{\angle B-\angle A}{2}=\frac{\angle A}{2}=\angle B A W$$

so the points $A, B, Y, W$ are concyclic, and therefore $Q A \cdot Q B=Q Y \cdot Q W$; hence, $Q$ has equal power with respect to $\Omega$ and $\omega$. It can be seen similarly that $R$ has equal power with respect to the circles $\Omega$ and $\omega$. Hence, the radical axis of $\Omega$ and $\omega$ is line $Q R$.

Let lines $O P$ and $Q R$ meet at point $S$. It is well-known that with respect to circle $A B C D$, the diagonal triangle $P Q R$ is autopolar. As consequences we have $O P \perp Q R$, and the points $P$ and $S$ are symmetric to circle $A B C D$, so $O S \cdot O P=r^{2}$.

Notice that $P$ and $O$ lie inside $\Omega$, so the polar line $Q R$ lies entirely outside, so $S$ is different from $P$ and $O$. Moreover,

$$S O \cdot S P=O S \cdot(O S-O P)=O S^{2}-O S \cdot O P=S O^{2}-r^{2}$$

so $S O \cdot S P$ is equal to the power of $S$ with respect to $\Omega$. Since $S$ lies on the radical axis $Q R$, it has equal power with respect to the two circles; therefore, $S O \cdot S P$ is equal to the power of $S$ with respect to $\omega$. From this, it follows that $O \in \omega$ if and only if $P \in \omega$.