Let $\angle B C M=\alpha$. Then

$$
\angle B M C=90^{\circ}-\alpha, \angle A M E=90^{\circ}-\left(90^{\circ}-\alpha\right)=\alpha, \angle A E M=90^{\circ}-\alpha .
$$

Points $B$ and $P$ view the segment $C M$ at a right angle, so these points lie on the circle with diameter $C M$. The inscribed angles $B P M$ and $B C M$ subtend the same arc in this circle, hence $\angle B P M=\angle B C M=\alpha$. Similarly, we can prove that $\angle A P M=90^{\circ}-\alpha$. Therefore,

$$
\angle A P B=\angle A P M+\angle B P M=\left(90^{\circ}-\alpha\right)+\alpha=90^{\circ} .
$$

## Answer

$90^{\circ}$.
4 [ Angles subtended by equal arcs and equal chords]

On side $A B$ of rectangle $A B C D$, a point $M$ is chosen. Through this point, a perpendicular to line $C M$ is drawn, which intersects side $A D$ at point $E$. Point $P$ is the foot of the perpendicular dropped from point $M$ to line $C E$. Find the angle $A P B$.