Solution:

(d) 100°

Explanation:

From triangles APB and CPD,

$\angle APB\text{ }=~\angle CPD\text{ }=\text{ }50{}^\circ $ (as they are vertically opposite angles)

$AP/PD\text{ }=\text{ }6/5$ … (i)

Also, $BP/CP\text{ }=\text{ }3/2.5$

Or $BP/CP\text{ }=\text{ }6/5$ … (ii)

From eq.(i) and eq.(ii),

We obtain,

AP/PD = BP/CP

As a result, ∆APB ∼ ∆DPC [using the SAS similarity criterion]

Therefore, ∠A = ∠D = 30° [as, corresponding angles of similar triangles are equal]

As we know that, the sum of angles of a triangle = 180°,

In triangle APB,

$\angle A\text{ }+~\angle B\text{ }+~\angle APB\text{ }=\text{ }180{}^\circ $

So, $30{}^\circ \text{ }+~\angle B\text{ }+\text{ }50{}^\circ \text{ }=\text{ }180{}^\circ $

Then, $\angle B\text{ }=\text{ }180{}^\circ (50{}^\circ +\text{ }30{}^\circ )$

$\angle B\text{ }=\text{ }18080{}^\circ =\text{ }100{}^\circ $

As a result, $\angle PBA\text{ }=\text{ }100{}^\circ $