Counter-example:

<b>Example</b>
  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
  $\forall A \in \operatorname{up} a : f \cap \prod A \ne \emptyset$.

<b>Proof</b>
  Take $\mathcal{L}_0$, $\mathcal{L}_1$ and $f$ from the the answer
  to <a href="http://mathoverflow.net/questions/189424/interweaving-two-indexed-families-of-filters">this question</a>.
  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 $a \not\in \upuparrows f$.