Problem 1369

1. **Reflecting Point $A$:**
Let $A'$ be the reflection of $A$ in the perpendicular bisector of $CE$. This implies that $A'$ lies on the line through $A$ perpendicular to $CE$ and equidistant from $C$ and $E$.

2. **Collinearity of Points $A', D, E$:**
Since $\angle CEA' = \angle ACE = \angle CED$, it follows that $A', D, E$ are collinear. This is because the reflection $A'$ maintains the angle properties with respect to $C$ and $E$.

3. **Similarity of Triangles $\triangle DEP$ and $\triangle PCA$:**
Given $\triangle DEP \sim \triangle PCA$, we have the proportionality of sides:
$$\frac{DE}{PC} = \frac{EP}{CA}$$
and the equality of angles:
$$\angle DEP = \angle PCA \quad \text{and} \quad \angle EDP = \angle ACP.$$

4. Equality of Lengths:
Since $PE = PC$, we can write:
$$EP^2 = EP \cdot PC = ED \cdot CA.$$

5. Circumcircle Radius Relation:
Let $R$ be the radius of the circumcircle $\odot (ADA')$ and $O$ be the circumcenter of $\triangle ADA'$. From the similarity and the given conditions, we have:
$$EP^2 = EO^2 - R^2.$$

6. Perpendicularity and Power of a Point:
Since $OP \perp BE$, we can use the power of a point theorem:
$$BO^2 - BP^2 = EO^2 - EP^2 = R^2.$$

7. Tangency Condition:
Using the power of a point theorem again, we get:
$$BD \cdot BA = BO^2 - R^2 = BP^2.$$
This implies that $BP$ is tangent to the circumcircle $\odot (PAD)$ at $P$.

$\blacksquare$