1. Let $DEF$ be the pedal triangle of point $P$ with respect to triangle $ABC$. The pedal triangle is formed by dropping perpendiculars from $P$ to the sides $BC$, $CA$, and $AB$, and letting the feet of these perpendiculars be $D$, $E$, and $F$ respectively.

2. Given the condition:
$$\angle BPC - \angle BAC = \angle CPA - \angle CBA = \angle APB - \angle ACB$$
we need to show that $DEF$ is an equilateral triangle.

3. Consider the angles:
$$\angle EDF = \angle PBA + \angle PCA = \angle BPC - \angle PAC$$
Similarly, we can express:
$$\angle DEF = \angle PCA + \angle APB = \angle CPA - \angle PBA$$
and
$$\angle EFD = \angle APB + \angle PBA = \angle APB - \angle PCA$$

4. From the given conditions, we have:
$$\angle BPC - \angle BAC = \angle CPA - \angle CBA = \angle APB - \angle ACB$$
This implies that:
$$\angle EDF = \angle DEF = \angle EFD$$
Therefore, triangle $DEF$ is equilateral.

5. Since $DEF$ is equilateral, point $P$ is known as the first isodynamic point of triangle $ABC$. By the properties of the isodynamic point, we have:
$$PA \cdot BC = PB \cdot CA = PC \cdot AB$$

6. This completes the proof that:
$$PA \cdot BC = PB \cdot CA = PC \cdot AB$$

$\blacksquare$