In figure, two line segments AC and BD intersect

(d) $\ln \triangle A P B$ and $\triangle C P D$,  $\angle A P B=\angle C P D=50^{\circ}$ [vertically opposite angles]

$\frac{A P}{P D}=\frac{6}{5}$ ...(i)

and $\frac{B P}{C P}=\frac{3}{2.5}=\frac{6}{5}$ ...(ii)

From Eqs. (i) and (ii) 

$\frac{A P}{P D}=\frac{B P}{C P}$

$\therefore$ $\triangle A P B \sim \triangle D P C$ [by SAS similarity criterion]

$\therefore$ $\angle A=\angle D=30^{\circ}$ [corresponding angles of similar triangles]

In $\triangle A P B, \quad \angle A+\angle B+\angle A P B=180^{\circ} \quad$ [sum of angles of a triangle $=180^{\circ}$ ]

$\Rightarrow \quad 30^{\circ}+\angle B+50^{\circ}=180^{\circ}$

$\therefore \quad \angle B=180^{\circ}-\left(50^{\circ}+30^{\circ}\right)=100^{\circ}$

i.e., $\quad \angle P B A=100^{\circ}$