Could I say that $\left(\mathbb{Z}/p^k\mathbb{Z}\right)_{b}$, namely the localization at $S=\begin{Bmatrix}b^n : n\in \mathbb{N}\end{Bmatrix}$ when $(b,p)=1$, is equal to $\mathbb{Z}/p^k\mathbb{Z}$ because we have that $$\left(\mathbb{Z}/p^k\mathbb{Z}\right)_{b}\cong\mathbb{Z}/p^k\mathbb{Z}\otimes \mathbb{Z}_{ b}$$ and there we have (taking $r=\max(n,k)$) $$ \bar{a}\otimes\frac{c}{b^n}= \bar{a}\otimes\frac{c}{b^n}\cdot 1=\bar{a}\otimes\frac{c}{b^n}\cdot(sb^r+tp^r)=\bar{a}\otimes csb^{r-n}+p^r\bar{a}\otimes \frac{ct}{b^n}=(csb^{r-n})\bar{a}\otimes 1 \ \ ?$$
I know that it is not a rigorous proof, but what I would like to say is that we can write every element of $\mathbb{Z}/p^k\mathbb{Z}\otimes \mathbb{Z}_{b}$ as an element of $\mathbb{Z}/p^k\mathbb{Z}$.
I would make it a little simpler, starting from a Bézout relation between $b$ and $p^k$, $\;ub+vp^k=1$. There results that for any $n$, $u^n$ is the inverse of $b^n$ modulo $p^k$. So an element in $\bigl(\mathbf Z/p^k\mathbf Z)_b$ can be written as $$\frac{\bar a}{b^n}=\frac{u^n\bar a}{u^nb^n}=\frac{u^n\bar a}1.$$