Diagonal AC and BD intersect at point O.
The following certificates PE+PF=DM= 12/5.
Because PE vertical AC, DM vertical AC, PN vertical DM,
Therefore, angle PEM= angle DME= angle PNM=90 degrees,
Therefore, the quadrilateral PEMN is a rectangle,
Therefore, PE=MN, PN and AC are connected in parallel, therefore, angle DPN= angle DAC.
Because the quadrilateral ABCD is a rectangle, AC=BD, and then AO=DO,
So, angle DAC= angle ADB, so, angle DPN= angle ADB.
In triangular PDN and triangular DFP
Angle PND= angle DFP=90 degrees, angle DPN= angle ADB, PD=DP,
Therefore, the triangular PDN congruent triangles DFP,
So, DN=PF.
Therefore, PE+PF=NM+DN=DM.
In right triangle ACD, AD=4, CD=3, then AC=5.
AD*CD=AC*DM,
So DM= 12/5, so there is PE+PF= 12/5.
(hope to adopt. )