Solution: AB=AC=4, the area of △ABC is 6, P is a point on BC, PD⊥AB is in D, PE⊥AC is in E.

Connecting AP, the area of △ABC = the area of △ APB+the area of △ APC.

∴ AB*PD/2 +AC*PE/2=6

AB(PD+PE)= 12

∴ PD+PE=3

∴ The value of PD+PE is certain.