Problem 1626

1. **Fixing Point $C$ on the Circle:**
We start by fixing point $C$ on the circle with chord $AB$. We need to show that $P$ is the reflection of $B$ across $AH$.

2. **Proving $P$ is the Reflection of $B$ across $AH$:**
Since $AH \perp BP$, we need to prove that $\measuredangle BAH = \measuredangle HAP$.

3. Introducing Perpendiculars:
Let $D$ and $F$ be the feet of the perpendiculars from $A$ and $C$ to $BC$ and $AB$ respectively. Since $\measuredangle AFC = \measuredangle ADC = 90^{\circ}$, quadrilateral $AFDC$ is cyclic.

4. Using Cyclic Quadrilateral Properties:
From the cyclic nature of $AFDC$, we have:
$$\measuredangle BAH = \measuredangle FAD = \measuredangle FCD = \measuredangle HCP = \measuredangle HAP$$
This shows that $P$ is the reflection of $B$ across $AH$.

5. **Extending $PH$ to Meet $(AHB)$ Again:**
Extend $PH$ until it meets the circumcircle of $\triangle AHB$ for a second time at point $X$. We will show that $X$ is a fixed point.

6. **Claim 1: $XB$ is Tangent to $(ABC)$:**
To prove this, we need to show that $\measuredangle XBA = \measuredangle BCA$.
$$\measuredangle XBA = \measuredangle XHA = \measuredangle PHA = \measuredangle PCA = \measuredangle BCA$$
This proves that $XB$ is tangent to $(ABC)$.

7. **Claim 2: $AX = AB$:**
We previously proved that $\measuredangle XBA = \measuredangle BCA$. Also, we have:
$$\measuredangle AXB = \measuredangle AHB = \measuredangle PHA = \measuredangle PCA = \measuredangle BCA$$
Since $\measuredangle AXB = \measuredangle XBA$, it follows that $AX = AB$.

8. **Defining $X$ Independently of $C$:**
Now, we can define $X$ based only on $A$, $B$, and the circle. $X$ is the intersection of the tangent at $B$ to the circle containing $A$, $B$, and $C$, and the circle centered at $A$ with radius $AB$ other than $B$. This point $X$ is independent of the position of $C$.

$$\boxed{X}$$