Let $b\in B$ be any point, and let $e_1, e_2, \ldots, e_k$ be the $k$ distinct elements of $p^{-1}(b_0)$ in $E$. Since $p$ is a covering map, for each $i=1,2,\ldots,k$, there exists an open neighborhood $U_i$ of $b$ in $B$ such that $p^{-1}(U_i)$ is a disjoint union of open sets $V_{i,1}, V_{i,2},\ldots,V_{i,k}$, each homeomorphic to $U_i$ via $p$. Note that $V_{i,j}$ contains exactly one element of $p^{-1}(b)$ for each $j=1,2,\ldots,k$, since the sets $V_{i,j}$ are disjoint.

Now let $e$ be any element of $p^{-1}(b)$. Since $B$ is connected, there exists a path $\gamma : [0,1] \rightarrow B$ from $b_0$ to $b$ in $B$. By the path lifting property, there exists a unique lift $\tilde{\gamma} : [0,1] \rightarrow E$ of $\gamma$ such that $\tilde{\gamma}(0) = e_1$. Since $p^{-1}(b_0)$ has $k$ elements, there exists a permutation $\sigma$ of ${1,2,\ldots,k}$ such that $\tilde{\gamma}(1) = e_{\sigma(1)}$. Moreover, since $\tilde{\gamma}(1)$ lies in the open set $V_{\sigma(1),j}$ for some $j=1,2,\ldots,k$, we have $p(e) = p(\tilde{\gamma}(1)) = p(e_{\sigma(1)})$, which implies that $e$ also lies in the open set $V_{\sigma(1),j}$ and therefore corresponds to the unique element $e_{\sigma(1),j} \in p^{-1}(b)$ in this open set.

Thus, we have shown that every element of $p^{-1}(b)$ corresponds to a unique element of the form $e_{\sigma(1),j}$ for some permutation $\sigma$ of ${1,2,\ldots,k}$ and some $j=1,2,\ldots,k$. Therefore, $p^{-1}(b)$ also has $k$ elements. This shows that $E$ is a $k$-fold covering of $B$.