Solution: connect PE, PF and CF, and pass through point E and point F respectively, so that EG is perpendicular to AD in G.

So EGA angle EGF angle =90 degrees.

So the triangle EGF is a right triangle.

So ef 2 = eg 2+fg 2.

Because the rectangular ABCD is folded along EF

So CE=PE

CF=PF

AB=CD

AD = BC

Angle ABC= angle BAD= angle BCD= angle ADC=90 degrees.

AB parallel CD

Parallel BC

So angle ABC= angle PAD=90 degrees.

Angle EGF= angle BAD=90 degrees

So AB is parallel to EF.

So the quadrilateral ABEG is a parallelogram.

So AB=EG

BE=AG

In the right triangle EBP, the angle ABC=90 degrees.

So PE 2 = Pb 2+Be 2.

Because AB=4 AP=2

PB=AB+AP

So PB=6

EG=4

Because AD=8

So BC=8

So BE=BC-CE=8-PE.

So PE 2 = 6 2+(8-PE) 2.

So AG=BE=7/4.

In the right triangle PAF, the angle PAF=90 degrees.

So pf 2 = AP 2+AF 2.

In a right triangle CDF, the angle ADC=90 degrees.

So cf 2 = df 2+CD 2

So AP 2+AF 2 = CD 2+DF 2.

Because DF=AD-AF=8-AF

AB=CD=4

So af 2+4 = 16+(8-af) 2.

In AF= 19/4.

Because FG=AF-AG=3

So EF= radical sign (3 2+4 2) = 5)

So EF=5