By sine law on $\bigtriangleup PKB$ and $\bigtriangleup PKC$,$\frac{CK}{\sin \angle PBK}=\frac{BP}{\sin \angle AKP}$,$\frac{CK}{\sin \angle PCK}=\frac{CP}{\sin \angle DKP} \implies \frac{CP}{BP}=\frac{\sin \angle PBK}{\sin \angle PCK}$,since $\sin \angle ABP=\sin \angle PBK,\sin \angle DCP=\sin \angle PCK,\frac{CP}{BP}=\frac{\sin \angle PBC}{\sin \angle PCB} \implies \frac{\sin \angle PBC}{\sin \angle PCB}=\frac{\sin \angle ABP}{\sin \angle DCP}$.Let $BP,CP$ intersect $AC,BD$ respectively at $M,N$.By sine law on $\bigtriangleup ABC$ and $\bigtriangleup CBD$,we deduce $\frac{AM}{CM}=\frac{AB}{BC} \times \frac{\sin \angle ABM}{\sin \angle CBM}=\frac{DN}{BN}=\frac{CD}{BC} \times \frac{\sin \angle DCN}{\sin \angle BCN}$,this along with our previous result implies $\frac{AB}{CD}=\frac{\sin \angle DCN}{\sin \angle ABM} \times \frac{\sin \angle CBM}{\sin \angle BCN}=1$ and we are done.