It suffices to show that $\angle TKH_1 = \angle TH_1B$ (Fig. 14). Let us extend $KH_1$ and intersect if with $w$ at $S$. Then
$$\angle TKS = \angle TAB + \angle BAS, \quad \angle TH_1B = \angle TAB + \angle PAC,$$
So it is sufficient to show that $\angle PAC = \angle BAS$.
Denote by $A_1$ the point such that $AA_1$ is the diameter of $w$. Then $\angle A_1CA = \angle ABA_1 = 90^\circ$, so $BH \parallel A_1C$ and $CH \parallel A_1B$. Then $BHCA_1$ is a parallelogram, so $HA_1$ passes through the point $M$.

Then $K$ lies on $HA_1$, hence $\angle A_1KA = 90^\circ$. Then the quadrilateral $AKH_1M$ is inscribed. So $\angle KH_1A = \angle KMA$. Suppose $AH_1$ intersect $w$ secondly at $F$. Then
$$\angle KH_1A = \angle KCA + \angle FAS, \quad \angle KMA = \angle KCA + \angle PAA_1.$$
Hence, $\angle FAS = \angle PAA_1$, also $\angle ABC = \angle AA_1C$, so we derive $\angle BAF = 90^\circ - \angle ABC = \angle A_1AC$. Then
$$\angle BAS = \angle BAF + \angle FAS = \angle PAA_1 + \angle A_1AC = \angle PAC.$$