Let angle ABC = 2x. Then ABP = PBC = x. And PAB = 180-2x.

In triangle PAB, Angle APB = 180 - PAB - ABP = x. So triangle PAB is isosceles, with base BP=6.

Since BP=PC=6, triangle BPC is also isosceles with base BC, and sides of length 6. So angle BCP = x, and angle BPC = 180-2x.

By similar angles, triangles ABP and BPC are similar. The following ratio leads to the solution (Note that AB=AP, and AD=BC):

  AB        6
-----  =  -----   ==>   (AB)^2 + 5(AB) - 36 = 0
  6        AB+5

Refer to the following, badly drawn, ascii diagram for reference:

                   B --------------- C
                     \ `           '\
                      \  6`      '6  \
                       \     ` '   5  \
                      A ------.-------- D
                              P

Given: BP bisects angle ABC
       PD=5, BP=6, and CP=6.
Find AB

1)  Since BP bisects angle ABC, angle ABP = angle PBC.
2)  Since BP = CP, anlge PBC = angle BCP.
3)  By opposite inner angles, angle CPD = angle BCP.
3a) Let Theta = ABP = PBC = BCP = CPD
4)  Sum of angles in triangle ABP: ABP + BPA + PAB = 180.
      4a) Theta + BPA + PAB = 180
5)  Sum of angles on line APD:     BPA + BPC + CPD = 180.
      5a) Theta + BPA + BPC = 180
6)  Eqn 4a) minus Eqn 5a) ==> PAB - BPC = 0
      6a) PAB = BPC
7)  Since angle PAB = angle BPC and angle ABP = angle PBC, 
    angle PBA = angle BCP, therefore triangle ABP is similar
    to triangle PBC.
      7a) AB/BP = BP/BC
      7b) Triangle ABP is isosceles w/ AB = AP
8)  AD = AP + PD, and AD = BC, therefore BC = AB + PD
9)  Substituting 8) and given lengths into 7a),
      AB/6 = 6/(AB+5)
      AB*(AB + 5) = 36
      AB^2 + 5*AB - 36 = 0
      (AB + 9)*(AB - 4) = 0
      AB = -9 or 4

      Choosing positive value gives AB = 4.