Let $A \subset \mathcal{L}(\mathcal{H}_0)$ be a concrete $C^*$-algebra and $X$ be a right Hilbert A-module. For each $x,y \in X$ we have rank one operators \begin{align} \theta_{x,y}: X & \to X \\ z & \mapsto x\langle y, z\rangle_A \end{align} The set $\mathcal{K}_A(X):=\overline{\mathrm{span}\{\theta_{x,y} : x,y \in X\}}$ is well known to be a $C^*$ algebra.
Question: Can I always find a representation of $\mathcal{K}_A(X)$ on some Hilbert space $\mathcal{H}_1$ such that for any $x \in X$ the map $\theta_{x,x}$, regarded as an element of $ \mathcal{L}(\mathcal{H}_1)$, "factors trhough $\mathcal{H}_0$"? To be precise, can I always find an element $b_x \in \mathcal{L}(\mathcal{H}_1, \mathcal{H}_0)$ such that $\theta_{x,x}=b_x^* b_x$?
I can't seem to find this in any of my usual references for Hilbert modules. It seems to be true for all the examples I've tried. For instance, for a usual Hilbert space regarded as a right Hilbert $\mathbb{C}$-module, we have $b_x=\langle x|$ so that $b_x^*$ is multiplication by $x$.
However, I can't quite prove it in general nor find a counterexample. I don't mind assuming that $X$ is full, but I am not sure this is needed. Also, it might be that this only works for a carefully chosen $\mathcal{H}_1$, but this is fine as I care for a "concrete" result rather than a necessarily abstract one.
Any reference or help to prove/disprove the claim will be much appreciated.
Given $\mathcal{H}_0$ define $\mathcal{H}_1 := X \otimes_{A} \mathcal{H}_0$, which is a Hilbert space carrying a representation of $\mathcal{K}(X)$ given by $T(x \otimes h) = (Tx) \otimes h$. Define $b_x$ to be the "creation" operator $b_x \in \mathcal{L}(\mathcal{H}_0,\mathcal{H}_1)$ satisfying $b_x(h) = x \otimes h$ and then $b_x^*$ is the corresponding "annihilation" operator $b_x^*(y \otimes h) = \pi((x \mid y)_A) h$, where $\pi$ is the representation of $A$ on $\mathcal{H}_0$. Then $$ b_x b_x^* (y \otimes h) = x \otimes \pi((x \mid y)_A)h = x(x \mid y)_A \otimes h = \Theta_{x,x}(y) \otimes h. $$ Of course there's some checking to be done to make sure everything is well-defined.
I should also mention that the representation of $\mathcal{K}(X)$ on $\mathcal{H}_1$ is faithful since $\pi$ is faithful.