Solution:

There are two cases to consider.

First, assume that the lines $AD$ and $BC$ are not parallel and meet at $S$. Since $ABCD$ is cyclic, $\triangle ASB$ and $\triangle CSD$ are similar. Then, $AE / AB = CF / CD$ and $AE / CF = AB / CD = AS / CS$, so that $\triangle ASE$ and $\triangle CSF$ are also similar ($\angle SAE = \angle SCD$), and therefore, $\angle DSE = \angle CSF$.

By similarity, we have
$$\frac{SE}{SF} = \frac{SA}{SC} = \frac{AB}{CD} = \frac{PE}{PF}$$
which means that $SP$ is the bisector of angle $S$ in $\triangle FSE$. This implies that $\angle ESP = \angle FSP$ and hence $\angle ASP = \angle BSP$, so $SP$ is also the bisector of angle $S$ in $\triangle ASB$. This means that $P$ is equidistant from the lines $AD$ and $BC$. Thus

$$\frac{[APD]}{[BPC]} = \frac{AD}{BD}$$

which is a constant (we use the notation $[ABC]$ for the area of $\triangle ABC$).

For the second case, assume that $AD$ and $BC$ are parallel. Then $ABCD$ is an isosceles trapezoid with $AB = CD$, and we have $BE = DF$. Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. Then $ME = NF$ and $E$ and $F$ are equidistant from the line $MN$. Thus $P$, the midpoint of $EF$, lies on $MN$. Thus $P$ is equidistant from $AD$ and $BC$, and hence

$$\frac{[APD]}{[BPC]} = \frac{AD}{BD}.$$

which is a constant.

Solution:

There are two cases to consider. First, assume that the lines $AD$ and $BC$ are not parallel and meet at $S$. Since $ABCD$ is cyclic, $\triangle ASB$ and $\triangle CSD$ are similar. Since $AE / AB = CF / CD$, then $AE / CF = AB / CD = AS / CS$, and $\triangle ASE$ and $\triangle CSF$ are also similar $(\angle SAE = \angle SCD)$. Therefore, $\angle DSE = \angle CSF$.
By similarity, we have
$$\frac{SE}{SF} = \frac{SA}{SC} = \frac{AB}{CD} = \frac{PE}{PF},$$
which means that $SP$ is the bisector of angle $S$ in $\triangle FSE$. This implies that $\angle ESP = \angle FSP$ and hence $\angle ASP = \angle BSP$, so $SP$ is also the bisector of angle $S$ in $\triangle ASB$. This means that $P$ is equidistant from the lines $AD$ and $BC$. Thus
$$[APD]/[BPC] = AD/BC,$$
which is a constant (we use the notation $[ABC]$ for the area of $\triangle ABC$).

For the second case, assume that $AD$ and $BC$ are parallel. Then $ABCD$ is an isosceles trapezoid with $AB = CD$, and we have $BE = DF$. Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. Then $ME = NF$ and $E$ and $F$ are equidistant from the line $MN$. Thus $P$, the midpoint of $EF$, lies on $MN$. This implies that $P$ is equidistant from $AD$ and $BC$, and hence
$$[APD]/[BPC] = AD/BC.$$