A window with a semicircular top

Perimeter. The frame consists of the two vertical sides, the bottom side of length $2r$, and the semicircular arc of length $\pi r$ (the diameter of the semicircle is not part of the frame). So $2h+2r+\pi r=P$, hence $h=\dfrac{P-(2+\pi)r}2$, and $h>0$ requires $0<r<\frac P{2+\pi}$.

Area.
$$A(r)=2rh+\frac{\pi r^2}2=Pr-(2+\pi)r^2+\frac{\pi r^2}2=Pr-\frac{4+\pi}2\,r^2.$$

Maximum. $A'(r)=P-(4+\pi)r=0\iff r=\dfrac P{4+\pi}$, which lies in $\left(0,\frac P{2+\pi}\right)$. The parabola is concave, so this is the maximum, and
$$A_{\max}=\frac{P^2}{2(4+\pi)}.$$

Shape. Then $h=\dfrac{P-(2+\pi)r}2=\dfrac{(4+\pi)-(2+\pi)}{2}\,r=r$.

Answer. The height of the rectangular part equals the radius of the semicircle.