In fact being able to express $B$ in that form and saying that $B$ is recursively enumerable are equivalent.
Here's a proof using Kleene Normal Form when $P$ is any partial recursive relation. First I'll show that if $B\subseteq\mathbb{N}^n$ is r.e., then there is a recursive relation $P\subseteq\mathbb{N}^{n+1}$ such that $$B(x_1,\ldots, x_n)\iff\exists y P(x_1,\ldots, x_n,y).$$
If $B=\emptyset$ the result is clear. Else let $f:\mathbb{N}\to\mathbb{N}^n$ be a total recursive function such that $B=\operatorname{range}(f)$. Then we have \begin{align*}B(x_1,\ldots, x_n) &\iff\exists y (f(y) = x_1,\ldots, x_n) \\ &\iff\exists y\forall i\le n (f_i(y) = x_i)\end{align*} which is in the desired form.
Now we show that there is a partial recursive function $f:\mathbb{N}^n\to\mathbb{N}$ such that $B=\operatorname{dom}(f)$. We can write $B(x_1,\ldots, x_n)\iff\exists y P(x_1,\ldots, x_n,y)$ where $P$ is a recursive relation. Then we simply take $$f(x_1,\ldots, x_n)=\mu y P(x_1,\ldots, x_n, y).$$ $f$ is recursive and $f(x_1,\ldots, x_n)\downarrow\iff B(x_1,\ldots, x_n)$.
Finally, we'll use the Kleene Normal form theorem get a primitive recursive function $f:\mathbb{N}\to\mathbb{N}^n$ such that $B=\operatorname{range}(f)$ or $B=\emptyset$. Indeed, write $B=\operatorname{dom}(f)$ for partial recursive $f$. Then, there are primitive recursive functions $U$ and $T$ and some $e\in\mathbb{N}$ such that $$f(x_1,\ldots, x_n)=U(\mu c T(e, \langle x_1,\ldots, x_n\rangle, c)).$$ Then $$B(x_1,\ldots, x_n)\iff f(x_1,\ldots, x_n)\downarrow\iff\exists c T(e,\langle x_1,\ldots, x_n\rangle, c),$$ which is in the form we wanted.
It is easy to see that the last condition implies the usual definition for r.e. sets, so we are done.
