Let $Y$ and $Z$ be the feet of the altitudes from $B$ and $C$ to $CA$ and $AB$, respectively. So $Y$ and $Z$ are on the circle with diameter $AH$. Notice that $AG \perp PG \Leftrightarrow AG \perp GH$, so $G$ is also on the circle with diameter $AH$. Claim: There exists a spiral similarity centered at $G$ taking $E$ to $Y$, $F$ to $Z$, and $T$ to $M$. Proof: First, $\measuredangle GEF = \measuredangle GAF = \measuredangle GAZ = \measuredangle GYZ$ and $\measuredangle FGE = \measuredangle FAE = \measuredangle FAY = \measuredangle ZAY = \measuredangle ZGY$. Thus $\triangle GEF \sim \triangle GYZ$. So there exists a spiral similarity centered at $G$ taking $E$ to $Y$ and $F$ to $Z$. From the Three Tangents Lemma, $MY$ and $MZ$ are tangent to the circumcircle of $GYZ$. Since $TE$ and $TF$ are also tangent to the circumcircle of $GEF$, the same spiral similarity also takes $T$ to $M$. This proves the claim. Because of the claim, we can deduce $\triangle GZF \sim \triangle GMT$, so to finish, $\measuredangle ALG = \measuredangle TMG = \measuredangle FZG = \measuredangle AZG$. So $AZLG$ is cyclic. But $(AZG)$ is the circle with diameter $AH$, so $AL \perp LH$ and we are done. $\blacksquare$