If $A$ is integrally closed in $K := \operatorname{Quot}(A)$ and $B$ is the integral closure of $A$ in a finite Galois extension $L$ over $K$, and there are prime ideals $\mathfrak{p}$ and $\mathfrak{P}$ of $A$ and $B$ respectively such that $\mathfrak{P}$ lies over $\mathfrak{p}$, when Lang writes that the decomposition group $Z_\mathfrak{P}$ of $\mathfrak{P}$ acts in a natural way on $B/\mathfrak{P}$, does he mean that for some $b + \mathfrak{P} \in B/\mathfrak{P}$ and $\tilde{\sigma} \in Z_\mathfrak{P}$ the action is $\tilde{\sigma}(b + \mathfrak{P}) = \sigma(b) + \mathfrak{P}$, where $\sigma \in \operatorname{Gal}(L/K)$?
Although, having written this, I realise that $\sigma(b) = b$ for all $b \in B$ and $\sigma \in \operatorname{Gal}(L/K)$ since $B$ is the integral closure of $A$, so that $\sigma(B) = B$, so this is wrong...
What does Lang mean by "acts in a natural way on $B/\mathfrak{P}$"?
This is correct : if $\sigma \in \mathrm{Gal}(L/K)$, then $$\sigma \cdot (b + \mathfrak{P}) := \sigma(b) + \mathfrak{P}$$ provides an action of the decomposition subgroup $Z_{\frak P} \leq \mathrm{Gal}(L/K)$ on the quotient ring $B / \mathfrak{P}$.
It is well-defined: if $b + \mathfrak{P} = b' + \mathfrak{P}$ then $b-b' \in \mathfrak{P}$ so that $\sigma(b-b') \in \frak P$ and hence $\sigma(b) + \mathfrak{P} = \sigma(b') + \mathfrak{P}$.