Solution:

Since $\triangle ABC$ is isosceles with $AB=BC,$ we have $\angle A=\angle C.$ With $\angle A=\dfrac52\angle B,$ the angle sum gives $$\dfrac52\angle B+\dfrac52\angle B+\angle B=180^\circ,$$ so $\angle B=30^\circ$ and $\angle ACB=75^\circ.$

By vertical angles $\angle DCE=75^\circ.$ Since $CD=CE,$ triangle $CDE$ is isosceles, so $$2\angle D+75^\circ=180^\circ,$$ giving $\angle D=52.5^\circ.$

Thus, the correct answer is A.