As each $P_A$ is a product of at most $l$ elements of {$2,4,\dots,2^t$} counted with multiplicity, we have for any $\lambda>0$,
$$\sum_{A \in \mathcal{P}} P_A^\lambda\leq(2^\lambda+2^{2\lambda}+\cdots+2^{t\lambda})^l<(2^{(t+1)\lambda}/(2^\lambda-1))^l.$$
Assuming $\lambda$ and $l$ are fixed, we obtain
$$\sum_{A \in \mathcal{P}} P_A^\lambda\ll(2^{t+1})^{\lambda l}<(4m^{1/r})^{\lambda l}\ll m^{\lambda l/r}.$$
Applying this for $\lambda:=r/l$ yields, for fixed $r$ and $l$,
$$\sum_{A \in \mathcal{P}} P_A^{r/l} \ll m.$$

EDIT: I corrected an oversight in my original message.