How will you solve it?

Since $\Delta CDP$ is equilateral and $ABCD$ is square, $PC=CB$ . So $\Delta PCB$ is isosceles.

Also, $\angle PCB = 30^{\circ}$ ; so $\angle CPB = \boxed{75^{\circ}}$ .

Here's the solution

$In$ $square$ $ABCD$ $and$ $Equilateral$ $triangle$ $PCD$ .

Details

AC=CD=PD=BD

ANGLE(ACD)=45 DEGREE ANGLE(PDO)=15 DEGREE

FROM DETAIL 1 PB=BD SO ANGLE(P)=ANGLE(B) P=B=X

NOW we have get all informations IT'S time for solving

In triangle DBP WE say :2X+30=180

WE get our answer as *75 * degree

$\overline{PC} = \overline{BC} \quad \implies \quad \angle CPB = \boxed{75^{\circ}}$