$\newcommand{\al}{\alpha}
\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\varepsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\om}{\omega}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\F}{\mathcal{F}}
\newcommand{\E}{\operatorname{\mathsf E}}
\renewcommand{\P}{\operatorname{\mathsf P}}
\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}
\newcommand{\tf}{\widetilde{f}}$
The answer depends on whether the support (say $S$) of the distribution of $\xi_1$ contains $1$ or not. If $1\in S$, then the answer is no; otherwise, yes.
Indeed, suppose first that $1\in S$. For each natural $N$, let
\begin{equation}
A_N:=\Big\{\forall k \ge 0\ \prod_{i=1}^{N}\xi_{i+kN} > 6N\Big\}=\bigcap_{k=0}^\infty B_{N,k},
\end{equation}
where
\begin{equation}
B_{N,k}=\Big\{\prod_{i=1}^{N}\xi_{i+kN} > 6N\Big\}.
\end{equation}
The events $B_{N,0},B_{N,1},\dots$ are independent and for each $k$
\begin{align*}
\P(B_{N,k})=\P(B_{N,0})&\le
1-\P\big(\xi_i<(6N)^{1/N}\ \forall i=1,\dots,N\big) \\
&=1-\P\big(\xi_1<(6N)^{1/N}\big)^N=:q<1,
\end{align*}
since $1\in S$ and $(6N)^{1/N}>1$. So,
\begin{equation}
\P(A_N)=\prod_{k=0}^\infty \P(B_{N,k})\le \prod_{k=0}^\infty q=0,
\end{equation}
for each natural $N$. So,
\begin{equation}
P:=\P\Big( \exists N \in \mathbb{N} \text{ s.t. } \forall k \ge 0\ \prod_{i=1}^{N}\xi_{i+kN} > 6N \Big)
=\P\Big(\bigcup_{N \in \mathbb{N}} A_N\Big)=0\ne1.
\end{equation}
If now $1\notin S$, then $\P(\xi_i>c)=1$ for some real $c>1$ and hence almost surely $\prod_{i=1}^{N}\xi_{i+kN}>c^N > 6N$ for all $k$ and all large enough $N$. Hence, in this case $P=1$.
