Proof: As shown in Figure 1, construct $D H \perp A C$ at point $H$, and let $B I$ intersect $Y J$ at point $M$, and $I J$ intersect $A C$ at point $N$.
$$\begin{array}{l} \text { By } \angle M J I=\angle M B D=\frac{1}{2} \angle A B C-\angle C B D \\ =90^{\circ}-\angle C A D-\frac{1}{2} \angle A D C=\angle J D H \\ \Rightarrow I J / / D H \Rightarrow I J \perp A C . \end{array}$$

By the properties of the incircle,
$$\begin{array}{l} A N-N C=A B-B C=A D-D C \\ \Rightarrow A B+D C=A D+B C \end{array}$$
$\Rightarrow$ Quadrilateral $A B C D$ is a bicentric quadrilateral.
As shown in Figure 2, draw $Y_{1} X_{1} \perp A C$, intersecting $O D$ and $O B$ at points $Y_{1}$ and $X_{1}$, respectively.
By $I J / / X_{1} Y_{1}$, and $I, J, D, B$ being concyclic,
$\Rightarrow Y_{1}, D, X_{1}, B$ are concyclic
$$\begin{array}{l} \Rightarrow A P \cdot P C=P B \cdot P D=P X_{1} \cdot P Y_{1} \\ \Rightarrow A, X_{1}, C, Y_{1} \text { are concyclic } \\ \Rightarrow \frac{X_{1} P}{P Y_{1}}=\frac{\sin \angle X_{1} O P}{\sin \angle Y_{1} O P} \cdot \frac{O X_{1}}{O Y_{1}} . \\ \text { Also, } \frac{B P}{P D}=\frac{\sin \angle B O P}{\sin \angle D O P} \cdot \frac{B O}{O D}, \frac{O X_{1}}{O Y_{1}}=\frac{O D}{O B} \\ \Rightarrow \frac{X_{1} P}{P Y_{1}}=\frac{B P}{P D} \cdot \frac{D O^{2}}{O B^{2}} . \end{array}$$

Taking the tangency points $T$ and $S$ on $A D$ and $B C$, respectively. By Newton's theorem, $T, P, S$ are collinear.
$$\begin{array}{l} \text { By } \triangle O B S \backsim \triangle D O T \\ \Rightarrow \frac{B O}{D O}=\frac{B S}{O T}=\frac{O S}{D T} \Rightarrow \frac{B O^{2}}{O D^{2}}=\frac{B S}{D T} \\ \Rightarrow \frac{B S}{P B}=\frac{\sin \angle B P S}{\sin \angle P S B}=\frac{\sin \angle T P D}{\sin \angle P S C} \\ =\frac{\sin \angle T P D}{\sin \angle P T D}=\frac{T D}{P D} . \\ \text { Thus, } \frac{X_{1} P}{P Y_{1}}=1 . \end{array}$$

Therefore, $A C$ is the diameter of the circumcircle of quadrilateral $Y_{1} A X_{1} C$. Hence, points $X$ and $Y$ are symmetric with respect to $A C$.