jmc

The semi-circle with diameter BC has radius $\frac{1}{2}$ that of the semi-circle with diameter AB, and thus, has $\frac{1}{4}$ of the area. (Area of a circle $=\pi \times r^2$ - thus, if $r$ is half as large, that will be squared in the process). Therefore, the sum of their areas represents $\frac{5}{8}$ of a circle with diameter AB, and since the line CP splits this area exactly in half, that area would be $\frac{5}{16}$ of a circle with diameter AB. Therefore, the degree measure of that sector is $360\times \frac{5}{16}=\boxed{112.5}$