Counter-example:

Example There exists such an (infinite) set $N$ and $N$-ary relation $f$ that $\mathcal{P} \in \upuparrows f$ but there are no indexed family $a \in \prod_{i \in N} \operatorname{atoms} \mathcal{P}_i$ of atomic filter such that $\forall A \in \operatorname{up} a : f \cap \prod A \ne \emptyset$.

Proof Take $\mathcal{L}_0$, $\mathcal{L}_1$ and $f$ from the the answer to this question. Take $\mathcal{P} = \operatorname{up}\mathcal{L}_0 \cap \operatorname{up}\mathcal{L}_1$. If $a \in \prod_{i \in N} \operatorname{atoms} \mathcal{P}_i$ then there exists $c \in \{ 0, 1 \}^N$ such that $a_i \sqsubseteq \mathcal{L}_{c (i)} (i)$ (because $\mathcal{L}_{c (i)} (i) \ne 0$). Then from that example it follows that $(\lambda i \in N : \mathcal{L}_{c (i)} (i)) \not\in \upuparrows f$ and thus $a \not\in \upuparrows f$.

Counter-example:

Example There exists such an (infinite) set $N$ and $N$-ary relation $f$ that $\mathcal{P} \in \upuparrows f$ but there are no indexed family $a \in \prod_{i \in N} \operatorname{atoms} \mathcal{P}_i$ of atomic filter such that $\forall A \in \operatorname{up} a : f \cap \prod A \ne \emptyset$.

Proof Take $\mathcal{L}_0$, $\mathcal{L}_1$ and $f$ from the the answer to this question. Take $\mathcal{P} = \operatorname{up}\mathcal{L}_0 \cap \operatorname{up}\mathcal{L}_1$. If $a \in \prod_{i \in N} \operatorname{atoms} \mathcal{P}_i$ then there exists $c \in \{ 0, 1 \}^N$ such that $a_i \sqsubseteq \mathcal{L}_{c (i)} (i)$ (because $\mathcal{L}_{c (i)} (i) \ne 0$). Then from that example it follows that $(\lambda i \in N : \mathcal{L}_{c (i)} (i)) \not\in \upuparrows f$ and thus $a \not\in \upuparrows f$.