Let $U\subset B$ be evenly covered by $p:E\rightarrow B$ and $\{V_\alpha\}$ be the partition of $p^{-1}(U)$ into pieces. Number of elements of $\{V_\alpha\}$ is equal to the number of elements of $p^{-1}(b)$ for all $b\in U$.
Because each of $p|_{V_\alpha}$ is a homeomorphism, exactly one element of $p(b)$ exists in $V_\alpha$. No more, no less.
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