The internal and external angle bisectors divide the segment $AB$ internally and externally in the same ratio $|AC| : |BC|$. This can be seen, for example, with the aid of the Sine Rule, applied to the triangles $ACP$ and $BCP$ for the internal bisector and to the triangles $ACQ$ and $BCQ$ for the external angle bisector. In particular, we obtain
$$\frac{|AP|}{|BP|} = \frac{|AQ|}{|BQ|}.$$

Let $X$ and $Y$ be points on the line $AM$ such that $BX \parallel MQ$ and $BY \parallel MP$.

As $BX \parallel MQ$, we have $|AM|/|XM| = |AQ|/|BQ|$ and as $BY \parallel MP$, we have $|AM|/|YM| = |AP|/|BP|$. This implies $|AM|/|XM| = |AM|/|YM|$, hence $|XM| = |YM|$.
Because $M$ is on the circle with diameter $PQ$, $\angle PMQ = 90^\circ$ and so also $\angle YBX = 90^\circ$. Therefore, $B$ is on the circle with diameter $XY$. As $M$ was shown to be the midpoint of $XY$, we obtain $MB = XM = YM$. This implies now
$$\frac{|AP|}{|BP|} = \frac{|AM|}{|BM|}$$
If $P'$ is the intersection point of the angle bisector of $\angle AMB$ and $AB$, then
$$\frac{|AP'|}{|BP'|} = \frac{|AM|}{|BM|} = \frac{|AP|}{|BP|}$$
This implies that $P = P'$, and so $MP$ is the angle bisector of $\angle AMB$.