When I tried to deal with this problem, I defined 2 sets $$\begin{align*} C= \left\{ b \in B: |p^{-1}(b)| =k \right\} \\ D= \left\{ b \in B : |p^{-1}(b)| \neq k \right\} \end{align*}$$ Since $b_{0} \in C$ , so $C \neq \emptyset$ . Then I assume $D \neq \emptyset$, I want to find a contradiction by using $B$ is connected to prove $D$ is indeed empty.
I need to prove $C$ is open. I consider any $b \in C$. Suppose $p^{-1}(b)=\left\{ e_{1},\cdots,e_{k} \right\}$ are in $E$ . Then there exists a neighborhood $U$ such that $$\begin{align*} p^{-1}(U) = \bigsqcup_{n=1}^{k} V_{k} \end{align*}$$ Since $p \bigg|_{V_{n}}$ is homeomorphism, this ensures that there must be $k$ open sets $V_{k}$ which are mapped onto $U$.
Then, I want to prove that for each $b' \in U$ , $p^{-1}(b')$ still has exact $k$ preimages, and then we will have $$\begin{align*} U \subset C \end{align*}$$ which proves $C$ is open. However, I think that $b'$ must at least have $k$ preimages brought by $V_{1},\cdots, V_{k}$, but I can't say why it has exact $k$ preimages.
Suppose $|p^{-1}(b')|=k+1$, what will lead to a contradiction? I think the contradiction may relate to $b$ having exact $k$ preimages. Any help? Thanks.
Edit:
I'm wondering why the situation above won't happen. That is, there exists a $e_{k+1}'$ such that $$\begin{align*} p(e_{k+1}') = b' \end{align*}$$ which leads to $p^{-1}(b')$ has preimages more than $k$
You have the right idea, but it is not properly elaborated. You consider $b \in C$, but you should not write $p^{-1}(U) = \bigcup_{n=1}^k V_n$ because this presupposes that there are $k$ sheets over $U$.
All we know is that each point of $B$ has an evenly covered open neigborhood $U \subset B$ which means that $$p^{-1}(U) = \bigcup_{\alpha \in A} V_\alpha \tag{1}$$ with pairwise disjoint open $V_\alpha \subset E$ which are mapped by $p$ homeomorphically onto $U$. Let $p_\alpha : V_\alpha \to U$ denote the homeomorphism obtained by restricting $p$.
Given $x \in U$, the fiber $p^{-1}(x)$ has the form
$$p^{-1}(x) = \bigcup_{\alpha \in A} p_{\alpha}^{-1}(x) . \tag{2}$$
This formula seems to be the core issue in your question, so let us prove it carefully. Obviously each $p_{\alpha}^{-1}(x) \subset p^{-1}(x)$ since $p_{\alpha}$ is the restriction of $p$ to $V_\alpha$. Hence $p^{-1}(x) \supset \bigcup_{\alpha \in A} p_{\alpha}^{-1}(x)$. Conversely, let $y \in p^{-1}(x)$ (which means $p(y) = x)$. Since $p^{-1}(x) \subset p^{-1}(U)$, $(1)$ shows that there exists a unique $\alpha \in A$ such that $y \in V_\alpha$. We conclude $p_\alpha(y) = p(y) = x$, thus $y \in p_{\alpha}^{-1}(x)$. This shows $p^{-1}(x) \subset \bigcup_{\alpha \in A} p_{\alpha}^{-1}(x)$.
Since each $p_\alpha$ is a homeomorphism, each $E_{x,\alpha} = p_{\alpha}^{-1}(x)$ is a one-point subset of $V_\alpha$. The $E_{x,\alpha}$ are pairwise disjoint (because the $V_\alpha$ are pairwise disjoint), therefore the cardinality of $p^{-1}(x)$ has the same value $\lvert A \rvert$ for all $x \in U$.
But we know that the cardinality of $p^{-1}(b)$ with $b \in C$ is $k$, thus for all $x \in U$ the cardinality of $p^{-1}(x)$ is $k$. This proves $U \subset C$.
That $D$ is open is proved by the same argument. If $b \in D$, then $p^{-1}(b)$ has a cardinality $\aleph \ne k$ and thus all fibers $p^{-1}(x)$, where $x$ lies in an evenly covered open neighborhood $U$, have cardinality $\aleph \ne k$. Hence $U \subset D$.