Label the points as shown:

Because $AB=4$ and the square side $BC=5,$ right triangle $ABC$ gives $AC=3.$ Also $BC\perp CE$ and $A,C,D$ are collinear, so $\angle ABC=\angle DCE.$ The right triangles $ABC$ and $CDE$ have equal hypotenuses $BC=CE=5,$ so they are congruent. Thus $CD=4, DE=3,$ and $EF=4-DE=1.$

Right triangles $EFG$ and $CDE$ are similar, so $$\frac{EG}{EF}=\frac{EC}{CD}=\frac54.$$ Hence $EG=5/4.$ The shaded region $BCEG$ is a trapezoid whose parallel sides are $BC=5$ and $EG=5/4,$ and whose height is the perpendicular side $CE=5.$ Its area is $$\frac12\left(5+\frac54\right)5=\frac{125}{8}=15\frac58.$$

Thus, the answer is D .