I am confused about poset $\sigma$-centered. Assume that $2 \leq |J| \leq \aleph_{0} $.  Let $\mathbb{P}=\operatorname{Fn}(I,J)$ 
$\mathbb{P}=\operatorname{Fn}(I,J)$  is $\sigma$-centered iff $|I| \leq \mathcal{c}$ where $\mathcal{c}=2^{\aleph_{0}}$
a suggestion to show this please.
Added: $\operatorname{Fn}(I,J)$ is the set of finite partial functions from $I$ to $J$, with the order $p\preceq q$ iff $p\supseteq q$. A poset $\langle\Bbb P,\preceq\rangle$ is centred if for each finite $F\subseteq\Bbb P$ there is a $q\in\Bbb P$ such that $q\preceq p$ for each $p\in F$. A poset is $\sigma$-centred if it is the union of countably many centred subposets.
 A: You can use the Hewitt-Marczewski-Pondiczery theorem, which says that if $\kappa$ is an infinite cardinal, $\mathscr{X}$ is a family of at most $2^\kappa$ spaces, each of density at most $\kappa$, then $\prod\mathscr{X}$ has density at most $\kappa$. Here we give $J$ the discrete topology and take $\kappa=\omega$ to conclude that ${^JI}$, viewed as the product of $|J|$ copies of $I$, is a separable space. Let $D$ be a countable dense subset of ${^JI}$. 
For each $f\in\Bbb P$ the set $B(f)=\{x\in{^JI}:f\subseteq x\}$ is a basic open set in ${^JI}$, so it contains some member of $D$. For each $x\in D$ let 
$$\Bbb P_x=\{f\in\Bbb P:x\in B(f)\}=\{f\in\Bbb P:f\subseteq x\}\;;$$
then $\Bbb P=\bigcup_{x\in D}\Bbb P_x$, and it suffices to show that each $\Bbb P_x$ is centred, which is straightforward.
There’s a sketch of a proof of the $\kappa=\omega$ case of the Hewitt-Marczewski-Pondiczery theorem here; Hewitt’s original paper can be found here; and there’s a proof here at Ask a Topologist by Henno Brandsma.
