Why didn't angle bisector theorem work?

Since BP bisects the right angle at B, we can designate BC =x and AB =2x. By Pythagoras' theorem we have:

3^2 =x^2 + 2x^2

9 = 5x^2

x =sqrt(9/5)

The area of ABC triangle = 1/2 * x * 2x =1/2 * sqrt(9/5) * 2sqrt(9/5) =9/5 units^2.