Let $(\mathbb{Z}^d, \mathbf{E}^d)$ be a graph with vertex set $\mathbb{Z}^d$ and edge set $\mathbf{E}^d$, such that $\mathbf{E}^d = \{(x, y) \in \mathbb{Z}^d \times \mathbb{Z}^d : \sum_{i = 1}^{d} |x_i - y_i| = 1\}$.
Let $(\Omega, \mathcal{A})$ be a measurable space, such that $\Omega = \prod_{e \in \mathbf{E}^d}\{0, 1\}$ and $\mathcal{A} = \sigma(\text{cylinder sets})$, with a measure $\mu$ defined on it.
Let $(\bar{\Omega}, \bar{\mathcal{A}}, \mathbb{P}_p)$ be a measure space, such that $\bar{\Omega} = \prod_{x \in \mathbb{Z}^d}\{0, 1\}$, $\bar{\mathcal{A}} = \sigma(\text{cylinder sets})$ and $\mathbb{P}_p$ is the Bernoulli Product Measure. In such a space, let $(\xi_x)_{x \in \mathbb{Z}^d}$ be a sequence of i.i.d. random variables, such that $\mathbb{P}_p(\xi_x = 1) = p$ and $\mathbb{P}_p(\xi_x = 0) = 1 - p$.
In order to indirectly define $\mu$, for a fixed $k \in \mathbb{N}$, say that $\omega_e = 1$ (for some $\omega \in \Omega$) if there exists $x \in \mathbb{Z}^d$ such that $e \in \Lambda_k(x)$ and $\xi_x = 1$. Where $\Lambda_k(x):= [-k, k]^d + (x_1, \cdots, x_d)$; i.e., $\Lambda_k(x)$ is a box of side $2k$ centered in $x$.
How can one prove that $\mu(A) = \mathbb{P}_p(\bar{A})$ for every $\bar{A} \in \bar{\mathcal{A}}$ event that induces the occurrence of $A \in \mathcal{A}$?
I'll just write out our conclusion.
By definition if $X:(\Omega,P)\to \Psi$ is a measurable map from a measure space into a measurable space, we define its push-forward measure by $$ X^*P(A)=P(X^{-1}(A)) $$ We seem to have concluded that you're in the scenario where $(\Omega,P)=(\{0,1\}^{\mathbf{E}^d},\mathbb{P}_p)$ and $\Psi=\{0,1\}^{\mathbf{Z}^d}$ with their respective cylinder $\sigma$-algebras.
Then, you define $X:\Omega\to \Psi$ by $$ X(\omega)(x)=\max_{e\in \Lambda_k(x)} \omega(e) $$ and define $\mu=X^*\mathbb{P}_p$, i.e. as the distribution of $X$. Defining $\overline{A}=X^{-1}(A),$ i.e. the set of all configurations $\omega$ such that $X(\omega)\in A$, we now get by definition that $\mu(A)=\mathbb{P}_p(\overline{A})$.