A Hilbert-space completion of a Hilbert $C^{*}$-module over a separable $C^{*}$-algebra

Let $B$ be a separable $C^{*}$-algebra and $\mathcal{E}$ a Hilbert $B$-module. We know that $B$ has a faithful state $\phi$. Using $\phi$, we can construct a $\mathbb{C}$-valued pre-inner product $[\cdot,\cdot]$ on $B$ by $$\forall \xi,\eta \in \mathcal{E}: \quad [\xi,\eta] \stackrel{\text{df}}{=} \phi(\langle \xi,\eta \rangle_{B}),$$ where $\langle \cdot,\cdot \rangle_{B}$ denotes the $B$-valued inner product on $\mathcal{E}$.

Question: Is it necessarily true that $\mathcal{E}$ is already complete with respect to the metric induced by $[\cdot,\cdot]$? If the answer is ‘no’, then is there a judicious choice of a faithful state $\phi$ on $B$ that would make this true?

Thank you very much.

One example of such a bimodule $\mathcal E$ is $B$ itself, with the inner product $\langle b,b'\rangle_B = b^* b'$. Choose $B\supset B_0 = C(X)$ a unital abelian $*$-subalgebra, which can be identified with the algebra of bounded continuous functions on a compact space $X$. If what you are asking were true, you would get that $C(X)\subset B$ is the same as its $L^2$-completion $L^2(X,\mu)\subset L^2(B,\phi)$ for some measure $\mu$ which assigns non-zero values to each non-empty open subset of $X$. From this you see that with the choice $\mathcal E=B$, the state $\phi$ with the property you want exists iff $B$ is finite-dimensional.