Let $\omega_B$ and $\omega_C$ be the circles with diameters $BB'$ and $CC'$, respectively. Let $QR$ be a chord in $\omega_B$, such that $P\in QR$ and $QR\perp BB'$ and similarly, let $ST$ be a chord in $\omega_C$, such that $P\in ST$ and $ST\perp CC'$. By Thales' Theorem, $\angle BQB'=90^\circ$, so $BQB'$ is a right triangle with altitide $QP$, thus we get $PQ^2=PB\cdot PB'$. Since $P$ lies on the diameter $BB'$ of $\omega_B$, $P\in QR$ and $QR\perp BB'$, we get $PQ=\frac{1}{2}QR$. Using analogous arguments for $\omega_C$ and using the condition $QR=ST$, we get $PB\cdot PB'=PQ^2=\left(\frac{1}{2}QR\right)^2=\left(\frac{1}{2}ST\right)^2=PT^2=PC\cdot PC'$. Therefore, $BCB'C'$ is cyclic. Similarly, $CAC'A'$ is cyclic. Then, we get $\angle C'AP\equiv\angle C'AA'=\angle C'CA'\equiv\angle C'CB=\angle C'B'B\equiv\angle C'B'P$. Therefore, $C'AB'P$ is also cyclic. From this, we get that $\angle BPC=\angle B'PC'=180^\circ-\angle C'AB=180^\circ-\alpha.$. Similarly, $\angle CPA=180^\circ-\beta$ and $\angle APB=180^\circ-\gamma$, so $P\equiv H \blacksquare$

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Firstly note that by Power of Point at $P$ , $ABA'B$ is cyclic. ( and cyclic variants hold) . Note that $$\measuredangle{AA'B}=\measuredangle{AB'B}=\measuredangle{BB'C}=\measuredangle{BC'C}=\measuredangle{AC'C}=\measuredangle{AA'C}$$ Hence $AA'$ is a altitude . Same holds for $BB'$ and $CC'$ and hence the lines concur at the orthocenter. $\blacksquare$