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If $g=xh{x}^{-1}$ in $G$, applying each projection map gives $p_j(g)=p_j(x)p_j(h)p_j(x{)}^{-1}$ in $G_j$.

Conversely, suppose $p_j(g)$ and $p_j(h)$ are conjugate for every $j$. Define the nonempty finite setEvery transition map $G_j\to G_i$ carries $X_j$ into $X_i$, so the $X_j$ form an inverse system. By the nonemptiness theorem for inverse limits of finite sets, there is a compatible tuple $(x_j)\in {\underleftarrow{\lim }}_j{X}_j\subseteq G$. Coordinatewise equality then gives $xh{x}^{-1}=g$. This proves the finite-quotient criterion for conjugacy in a profinite group.

Suppose first that $\Gamma$ is conjugacy separable. If $h\in \Gamma$ is not conjugate to $\gamma$ in $\Gamma$, some homomorphism to a finite group sends them to nonconjugate elements. This homomorphism factors through a finite quotient of $\hat{\Gamma }$, so $h$ cannot be conjugate to $\gamma$ in $\hat{\Gamma }$. Thus

Conversely, suppose this equality holds and $h$ is not conjugate to $\gamma$ in $\Gamma$. Then they are not conjugate in $\hat{\Gamma }$. By the finite-quotient criterion for conjugacy in a profinite group, their images fail to be conjugate in some finite quotient. This is precisely conjugacy separability.