Solution:
WLOG assume $P$ is between $A$ and $B$, and let $s$ be the side length of $\triangle ABC$. By Ptolemy's Theorem,
$$(PA + PB) \cdot s = PA \cdot BC + PB \cdot AC = PC \cdot AB = PC \cdot s \Longrightarrow PA + PB = PC.$$
By the law of cosines on $\triangle PAB$,
$$PA^{2} + PB^{2} - 2 PA \cdot PB \cos \angle PAB = AB^{2}.$$
Since $\angle PAB = 120^{\circ}$ and $\cos 120^{\circ} = -\frac{1}{2}$, this becomes
$$PA^{2} + PB^{2} + PA \cdot PB = s^{2}.$$
Squaring both sides gives
$$PA^{4} + 2 PA^{3} \cdot PB + 3 PA^{2} \cdot PB^{2} + 2 PA \cdot PB^{3} + PB^{4} = s^{4}.$$
Doubling both sides and rearranging gives
$$\begin{aligned} 2s^{4} & = 2 PA^{4} + 4 PA^{3} \cdot PB + 6 PA^{2} \cdot PB^{2} + 4 PA \cdot PB^{3} + 2 PB^{4} \\ & = PA^{4} + PB^{4} + (PA + PB)^{4}. \end{aligned}$$
Using the fact that $PA + PB = PC$, we get
$$PA^{4} + PB^{4} + PC^{4} = 2s^{4}.$$
Thus, the sum does not depend on $P$.