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 $\lambda>0$, $$\sum_{A \in \mathcal{P}} P_A^\lambda\leq(2+4+\cdots+2^t)^{l\lambda}<2^{(t+1)l\lambda}<(4m^{1/r})^{l\lambda}.$$ Applying this for $\lambda:=r/l$ yields $$\sum_{A \in \mathcal{P}} P_A^{r/l} < 4^r m.$$

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.

As each $P_A$ is a product of the elements of {$2,4,\dots,2^t$}, each element repeated at most $l$ times, we have for any $\lambda>0$, $$\sum_{A \in \mathcal{P}} P_A^\lambda\leq(2+4+\cdots+2^t)^{l\lambda}<2^{(t+1)l\lambda}<(4m^{1/r})^{l\lambda}.$$ Applying this for $\lambda:=r/l$ yields $$\sum_{A \in \mathcal{P}} P_A^{r/l} < 4^r m.$$

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 $\lambda>0$, $$\sum_{A \in \mathcal{P}} P_A^\lambda\leq(2+4+\cdots+2^t)^{l\lambda}<2^{(t+1)l\lambda}<(4m^{1/r})^{l\lambda}.$$ Applying this for $\lambda:=r/l$ yields $$\sum_{A \in \mathcal{P}} P_A^{r/l} < 4^r m.$$