如图所示,设PF⊥CD,
∵BP=FP,
由翻折变换的性质可得BP=B′P,
∴FP=B′P,
∴FP⊥CD,
∴B′,F,P三点构不成三角形,
∴F,B′重合分别延长AE,DC相交于点G,
∵AB平行于CD,
∴∠BAG=∠AGC,
∵∠BAG=∠B′AG,AGC=∠B′AG,
∴GB′=AB′=AB=5,
∵PB′(PF)⊥CD,
∴PB′∥AD,
∴△ADG∽△PB′G,
∵Rt△ADB′中,AB′=10,AD=8,
∴DB′=6,DG=DB′+B′G=6+10=16,
∴△ADG与△PB′G的相似比为8:5,
∴AD:PB′=8:5,
∵AD=8,
∴PB′=5,即相等距离为5.
故答案为:5.