It is a question of synthetic geometry.Please solve it without using trigonometry.

this is an exceptionally challenging ques.

so you will have to apply internal angle bisector theorem on tri QAP. also extend AM to let it meet extended CD at R. from R draw a line parallel to BC and let it meet extended AB at G. apply similarity between tri AMB and tri CMR. further note that AP= PR (this observation is the heart of the proof), coz in tri APR angle RAP and angle PRA are both equal to PAB/2.

finally apply the formula for the length of internal angle bisector.

you will (after a hell lotta calculations) obtain 

p= (λ/a)(p^2 – b^2) – b

or p+b= (λ/a)(p – b)(p+b)

or p= a/λ+b