I'm trying to prove the question in the title, where the axioms are enunciated like
$FA_{\kappa}(\mathbb{P})$ is the assertion that for each colection $\{I_{\alpha} : \alpha < \kappa\}$ of maximal antichains of $\mathbb{P}$, there exists a filter $G\subseteq \mathbb{P}$ such that $I_{\alpha}\cap G \not = \emptyset$, for all $\alpha<\kappa$.
$BFA_{\kappa}(\mathbb{P})$ is the assertion that for each colection $\{J_{\alpha} : \alpha < \kappa\}$ of maximal antichains of $\ \mathbb{B}=RO(\mathbb{P})\setminus\{0\}$, with each one of size at most $\kappa$, there exists a filter $G\subseteq \mathbb{B}$ such that $J_{\alpha}\cap G \not = \emptyset$, for all $\alpha<\kappa$.
The only trick I'm working one is how to code antichanis and filters of $\mathbb{P}$ into antichains and filters of $\mathbb{B}$ (and vice versa).
My attempt on $\rightarrow$ part was, after assuming $FA_{\kappa}(\mathbb{P})$ and fixing an arbitrary antichain $J_{\alpha}=\{U_{\theta}\in \mathbb{B} : \theta < \kappa\}$ was to just pick one element of each regular open $U_{\theta}$ and produce an antichain of $\mathbb{P}$. But it turns out that the resulting antichain need not to be a maximal one.
My attempt on $\leftarrow$ part was, after assuming $BFA_{\kappa}(\mathbb{P})$ and fixing an arbitrary antichain $I_{\alpha}=\{p_{\theta}\in \mathbb{P} : \theta<\kappa\}$, I set $U_{p_{\theta}}=\{q\in\mathbb{P}: q\leq p_{\theta}\ \mbox{ or } \ p_{\theta}\leq q\} \in \mathbb{B}$. But the colection $J_{\alpha}$ made of these $U_{p_{\theta}}$ need not to be an antichain.
I also tried several other non-successful methods which I can exhibit if you judge useful.
Could you guys help me on this? Do I need to go to Stone Space arguments? To get a better characterization of the regular open sets or something like that...