Solution: the middle vertical line is AB,
∫△ABC is an equilateral triangle, △ABD is an isosceles triangle;
∴AB's middle vertical line must pass through C and D, ∠ BCE = 30;
AB = BP = BC,∠DBP=∠DBC,BD = BD
∴△BDC≌△BDP, so ∠ BPD = 30.
Therefore, the degree of ∠BPD is 30.
Note: This question is not difficult. The key to solve this problem is to make auxiliary lines and then use the properties of equilateral triangles to solve it.
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