Analysis: rotate △BPC counterclockwise around point B by 60 to get △BEA, and according to the nature of rotation, get BE=BP=4, AE=PC=5, ∠ PBE = 60, then △BPE is an equilateral triangle, and get PE=PB=4, ∠ BPE = 60, at △ AE.

Solution: Solution: ∫△ABC is an equilateral triangle,

∴BA=BC，

△BEA can be obtained by rotating △BPC 60 degrees counterclockwise around point B,

Even EP, as shown in the figure,

∴BE=BP=4,AE=PC=5,∠PBE=60，

∴△BPE is an equilateral triangle,

∴PE=PB=4,∠BPE=60，

At △AEP, AE=5, AP=3, PE=4,

∴AE2=PE2+PA2，

∴△APE is a right triangle, and∠ APE = 90,

∴∠APB=90 +60 = 150。

So the answer is 150.

Comments: This question examines the essence of rotation: the two figures are the same before and after rotation, the included angle between the corresponding point and the rotation center is equal to the rotation angle, and the distance between the corresponding point and the rotation center is equal. The judgment and properties of equilateral triangle and the inverse theorem of Pythagorean theorem are also investigated.