Question 147277
Let the angle at BAC = a. 
Triangle ABC is isosceles, so its base angles are equal. Call those angles b.
We have a+b+b = 180

By extending B to P, with a length equal to BC, we create a second isosceles triangle BPC. The angle at PBC is the supplement of the angle CBA. That angle is b (from above). So angle PBC = 180-b 
Since the second triangle also has base angles equal, we know that angle BPC and BCP are the same. Call that angle c.

Then PBC + 2c = 180.
PBC = 180 - 2c
From above PBC = 180-b, so
180-b = 180 - 2c
b/2 = c

The angle at CPB = c = b/2
The angle PCA = c + b = b/2 + b = 3b/2  
Which is 3 times CPB