To avoid config issues, we will use directed angles. We fix $C$ on the circle with chord $AB$. We will first show that $P$ is the reflection of $B$ across $AH$. $AH \perp BP$, so we just have to prove $\measuredangle BAH = \measuredangle HAP$. Let $D$ and $F$ be the feet of the perpendiculars from $A$ and $C$ to $BC$ and $AB$ respectively. Since $\measuredangle AFC = \measuredangle ADC = 90^{\circ}$, $AFDC$ is cyclic. So $\measuredangle BAH = \measuredangle FAD = \measuredangle FCD = \measuredangle HCP = \measuredangle HAP$. This shows $P$ is the reflection of $B$ across $AH$. Extend $PH$ until it meets $(AHB)$ for a second time at point $X$. We will show that $X$ is the fixed point. Claim 1: $XB$ is tangent to $(ABC)$. Proof: $\measuredangle XBA = \measuredangle XHA = \measuredangle PHA = \measuredangle PCA = \measuredangle BCA$. This proves that $XB$ is tangent to $(ABC)$. Claim 2: $AX = AB$. Proof: Previously, we proved $\measuredangle XBA = \measuredangle BCA$. Also, $\measuredangle AXB = \measuredangle AHB = \measuredangle PHA = \measuredangle PCA = \measuredangle BCA$ Since $\measuredangle AXB = \measuredangle XBA$, then $AX = AB$. Now we can define $X$ on just $A, B$, and the circle, independent of the position of $C$. $X$ is the intersection of the tangent at $B$ of the circle containing $A, B$, and $C $and circle centered at $A$ with radius $AB$ other than $B$. And we are done.

Let $\omega$ be circle symmetric to the given with respect to $AB$. It's a well-known fact that all points $H$ lie on $\omega$. Let $PS\cap \omega=S$. We'll prove that all lines $PH$ pass through the point $S$. $\angle BPH=\angle CAH=90^{\circ}-\angle C$; $\angle HBC=90^{\circ}-\angle C$. Then: $$\angle BHS=\angle HBC+\angle HPB=180^{\circ}-2\angle C$$this is constant, since $\angle ACB$ is constant. This means that point $S$ is same for all lines $PH$, meaning all lines $PH$ pass through point $S$

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