P is PE vertical AC at point E, PF vertical BD at point F, DM vertical AC at point M and PN vertical DM at point N. ..

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. )