Given that line PN is a perpendcular bisector of line AB, then angles APN and BPN are equal to 90 degrees each.
Also, given that line PN is an angle bisector to angle CPD and angle CPD measures x, then the measure of angle DPN is [tex] \frac{1}{2} x[/tex].
Thus, the measure of angle DPB = angle BPN - angle DPN
[tex]=90- \frac{1}{2} x[/tex]
Therefore,
[tex]\sin DPB=\sin\left(90- \frac{1}{2} x\right) \\ \\ =\sin(90)\cos\left(\frac{1}{2} x\right)-\sin\left(\frac{1}{2} x\right)\cos(90) \\ \\ =\cos(\frac{1}{2} x)[/tex]