Let 'D' be the center of the circle. So, points A,D,B,P all are on the same line. Also, P and C are points on the tangent.

Join CD and BC.

Now, ∠BCA is the angle inscribed in a semi-circle, so it's 90º.

C is the point on the circle where the tangent touches the circle. So, ∠DCP = 90º.

∠PCA =∠PCD + ∠DCA

110º = 90º + ∠DCA

So, ∠DCA =20º

In triangle ADC, AD = DC (Both are radii of the circle)

So, ∠DCA = ∠CAD =20º

In triangle ABC: ∠BCA = 90º, ∠CAB = 20º

So, ∠CBA = 70º.