Since AP=BP=AB, \triangle PAB is an equilateral triangle.

Since AC=BC, \triangle ABC is an isosceles triangle.

\triangle PAB and \triangle ABC share the same side AB, Point D is the middle point of AB. Connecting PD and CD, it can be found AB\perp PD and AB\perp CD .

Therefore, AB\perp plane PCD, which means AB is perpendicular to any line in the plane ABC, including line PC.

From (1), PC\perp AB and given condition PC\perp AC

PC is perpendicular to the plane ABC. Therefore,

PC\perp CD

\triangle PCD is a right triangle

Since \triangle ABC is an isosceles right triangle

AB = \sqrt{2} AC =2\sqrt{2}

PD = \cos 30\degree PB = \dfrac{\sqrt{3} }{2}\cdotp 2\sqrt{2} =\sqrt{6}

DC = \dfrac{AB}{2} =\sqrt{2}

Applying Pythagorean Theorem gives

PC = \sqrt{PD^2-CD^2}

=\sqrt{6-2} =2

Therefore, \triangle PAC is also an isosceles right triangle.

Find the midpoint of the segment PA.

In \triangle PAC ,

\because AC=PC. \therefore CE\perp PA

In \triangle PAB ,

\because PB=AB. \therefore BE\perp PA

Therefore, \angle BEC is the dihedral angle B−AP−C

CE = \dfrac{PA}{2}=\sqrt{2}

BE = PD = \sqrt{6}

BC=2

Using the Law of Cosines,

\cos \angle BEC = \dfrac{BE^2+CE^2-BC^2}{2\cdotp BE\cdotp CE }

=\dfrac{6+2-4}{2\cdotp \sqrt{6}\cdotp \sqrt{2} } =\dfrac{\sqrt{3} }{3}

The dihedral angle B−AP−C is \arccos\dfrac{\sqrt{3} }{3}