$$\boxed{\left\{\frac{4}{3},\ \frac{5}{3}\right\}}$$
We can draw, as in the figures below, a regular hexagon $ACEFGH$ which is inscribed in the circle given in the statement of the problem. As we are concerned with the absolute value of the difference between $BP$ and $DP$, the point $B$ is chosen to lie on the opposite side from the center of the circle with respect to the line $AC$.

Let us consider the case where $\angle APB = 60^\circ$ as indicated in the figure on the left side. Then, since $\angle PAF = 60^\circ$ also, the lines $BD$ and $AF$ are parallel. If we let $Q$ be the point of intersection of the lines $BD$ and $EF$, we have $BP = DQ$ due to the symmetry. Hence the absolute value of the difference between $BP$ and $DP$ equals $PQ$. From the fact that $AP : PC = FQ : QE = 1 : 2$ we obtain $PQ = \frac{5}{3}$ since $AF = 2$ and $CE = 1$.

In the case where $\angle APB = 120^\circ$ as indicated in the figure on the right side, let $Q$ be the point of intersection of $BD$ and $GH$. Then, similarly as in the preceding case, we deduce the fact that the absolute value of the difference between $BP$ and $DP$ equals $PQ$. From the fact that $AP : PC = HQ : QG = 1 : 2$ we obtain $PQ = \frac{4}{3}$ since $CG = 2$ and $AH = 1$.

Hence the possible values for the absolute value of the difference between $BP$ and $DP$ are $\frac{4}{3}, \frac{5}{3}$.