Let $AO$ meet $BC$ at $D$. Since
$$\angle DAC = \angle BAD = \frac{1}{2} \angle BAC = \angle ACD,$$
we have $DA = DC$. Consider $\triangle OCA$ and $\triangle OCD$. By considering their areas, we find that
$$\frac{AO}{OD} = \frac{[OCA]}{[OCD]} = \frac{AC \times d(O, AC)}{CD \times d(O, CD)} = \frac{AC}{CD}.$$

The last equality holds since $d(O, AC) = d(O, CD)$ is the A-exradius. Now, applying Menelaus' theorem to $\triangle ACD$, we obtain
$$\frac{DP}{PC} \times \frac{CM}{MA} \times \frac{AO}{OD} = 1.$$
Using $CM = MA$ and $\frac{AO}{OD} = \frac{AC}{CD} = \frac{AC}{AD}$, this becomes
$$\frac{PD}{PC} = \frac{AD}{AC}.$$

By the angle bisector theorem, we know that $AP$ bisects $\angle DAC$. Therefore, we get
$$\angle BAP = \angle BAD + \angle DAP = \angle ACP + \angle PAC = \angle APB.$$
Thus, $AB = BP$.