Hello, Kevin!
Draw radius $\displaystyle MQ$ downward, perpendicular to $\displaystyle AB.$
Let $\displaystyle \theta = \angle PMQ.$
Ignore the two smaller semicircles for the moment.
If the shaded region is the semicircle $\displaystyle APB$, then $\displaystyle MQ$ bisects the area.
But a semicircle with area $\displaystyle \tfrac{1}{2}\pi r^2$ has been subtracted from quadrant $\displaystyle MQB$
. . and added to quadrant $\displaystyle MQA.$
For the new area is to be bisected by radius $\displaystyle MP$,
. . the area of sector $\displaystyle PMQ$ must equal $\displaystyle \tfrac{1}{2}\pi r^2$
The area of sector $\displaystyle PMQ \:=\:\tfrac{1}{2}(2r)^2\theta \:=\:2r^2\theta$
So we have: .$\displaystyle 2r^2\theta \:=\:\tfrac{1}{2}\pi r^2 \quad\Rightarrow\quad \theta \:=\:\tfrac{\pi}{4}$
Hence: .$\displaystyle \angle PMQ \:=\:\angle AMP \:=\:\tfrac{\pi}{4}$
Therefore, we must draw radius $\displaystyle MP$ so that angle $\displaystyle AMP = 45^o.$