Find $B_n$ such that $\mathbb{P}_p(A \mathbin{\Delta} B_n)=0$ I have a question about percolation.

Show that any measurable event can be approximated by events depending on finitely many edges, in the sense that for any $A$ in the product $\sigma$-algebra, there exists a sequence $\{B_n\}$ such that $B_n$ is measurable in terms of the percolation configuration $\omega_e$ for $e$ in the box of size $n$, and $\mathbb{P}_p(A \mathbin{\Delta} B_n)=0$

where $\mathbb{P}_p$ is the product measure on $\{0,1\}^E$ if $\omega_e=1$ the edge $e$ is open with probability $p$.
How to find such $B_n$?
 A: You have not specified enough of the question for it to make sense, so I will fill in the details you have not specified and answer that question instead.
Consider Bernoulli percolation on $\mathbb Z^d$. We have a probability space $\Omega=\{0,1\}^{\mathbb Z^d}$ equipped with product $\sigma$-algebra and product measure $\mathbb P_p=\textrm{Ber}_p^{\mathbb Z^d}$. We have a sequence of i.i.d. random variables $(X_e)_{e\in\mathbb Z^d}$ distributed according to $\mathbb P_p$. For each $n\in\mathbb N$ we have a $\sigma$-algebra $\mathcal F_n=\sigma(X_e\colon e\in [-n,n]^d)$, i.e. the events that are determined by a box of size $n$.
I believe the question you are asking is to prove the following.
Lemma. For all measurable sets $A\subseteq \Omega$, there exists a sequence of sets $B_n\in\mathcal F_n$ such that $$\lim_{n\to\infty}\mathbb P_p(A\triangle B_n)=0.\qquad (\star)$$
Proof. Let $\mathcal F$ be the collection of all events that satisfy the property $(\star)$. I will start by showing that $\mathcal F$ is, in fact, a $\sigma$-algebra. It is clear that $\Omega\in\mathcal F$, since one can take the sets $B_n=\Omega\in\mathcal F_n$. Next, suppose that $A\in \mathcal F$ and $B_n$ approximates $A$ as in $(\star)$. Then $A^c$ is approximated by $B_n^c$, since by definition of the symmetric difference
$$
A\triangle B_n=(A\cap B_n^c)\cup (A^c\cap B_n)=A^c\triangle B_n^c.
$$
Hence $A^c\in\mathcal F$, so we have shown that $\mathcal F$ is closed under complementation. Finally, suppose that we have a sequence of sets $\{A_i\}_{i\in\mathbb N}$ with each $A_i\in\mathcal F$. To show that the countable intersection $A=\cap_i A_i$ is also in $\mathcal F$, observe that if for each $i$ we have an approximating sequence of sets $B_{in}$ for $A_i$, then the sets $B_n=\cap_i A_{in}$ form an approximating sequence for the intersection $A$, and therefore $A\in\mathcal F$. This completes the proof that $\mathcal F$ is a $\sigma$-algebra.
Now that we know $\mathcal F$ is a $\sigma$-algebra, we are nearly done. Indeed, we know that $\mathcal F$ is contained in the product $\sigma$-algebra for $\{0,1\}^{\mathbb Z^d}$. On the other hand, it is plain to see that for every finite collection of edges $E\subset \mathbb Z^d$, the $\sigma$-algebra $\sigma(X_e\colon e\in E)$ is contained in $\mathcal F$ - since for all $n$ sufficiently large, we will have $E\subseteq [-n,n]^d$ and then the approximation is exact for these large enough $n$. Since the product $\sigma$-algebra is generated by these "finite edge set" $\sigma$-algebras, it follows that $\mathcal F$ equals the entire $\sigma$-algebra, as desired. $\square$
