**Theorem:** For Hilbert spaces $H,K$, every normal unital *-homomorphism $\Phi:\mathcal B(H)\to\mathcal B(K)$ is of the form $\Phi(a)=U(a\otimes 1_{K_0})U^∗$ for some Hilbert space $K_0$ and some unitary $U$.

Here is a super down-to-earth proof, from a functional analysis / operator-algebras perspective. I'll start by working in the infinite-dimensional setting. Let $H,K$ be Hilbert spaces, write $\newcommand{\mc}{\mathcal}\mc B(H)$ for the algebra of all bounded operators on $H$, and $\mc K(H)$ for the compact operators. Let $\Phi:\mc B(H)\rightarrow\mc B(K)$ be a unital $*$-homomorphism which is *normal* (aka weak$^\ast$-continuous). As $\mc K(H)$ is weak$^\ast$-dense in $\mc B(H)$, $\Phi$ is completely determined by its restricton to $\mc K(H)$, That $\Phi$ is unital corresponds to $\Phi:\mc K(H)\rightarrow\mc B(K)$ being *non-degenerate* meaning that $\{ \Phi(\theta)(\xi) : \theta\in\mc K(H), \xi\in K \}$ has a dense linear span in $K$.

If $H$ is finite-dimensional, of course $\mc B(H) = \mc K(H) \cong M_i$ where $i$ is the dimension of $H$.

So, consider a non-degenerate $*$-homomorphism $\Phi:\mc K(H)\rightarrow\mc B(K)$. For $\xi,\eta\in H$ write $\theta_{\xi,\eta}$ for the rank-one operator $\alpha\mapsto (\alpha|\eta) \xi$. Then $\theta_{\xi,\eta}^* = \theta_{\eta,\xi}$ and $\theta_{\xi,\eta} \theta_{\xi_1,\eta_1} = (\xi_1|\eta) \theta_{\xi,\eta_1}$. Here I write $(\cdot|\cdot)$ for the inner-product on $H$. Fix a unit vector $\xi_0\in H$, and consider $\theta_{\xi_0, \xi_0}$ which is a projection (self-adjoint idempotent). So $p = \Phi(\theta_{\xi_0, \xi_0})$ is also a projection. Let $K_0\subseteq K$ be the (closed) subspace forming the image of $p$.

Define $U:H\odot K_0 \rightarrow K$ by
$$ U(\xi\otimes\alpha) = \Phi(\theta_{\xi,\xi_0})(\alpha), $$
and extend by linearity. I write $\odot$ for the algebraic tensor product, and will write $\otimes$ for the (completed) Hilbert space tensor product. Then
\begin{align*}
( U(\xi\otimes\alpha) | U(\eta\otimes\beta) )
&= (\Phi(\theta_{\xi,\xi_0})(\alpha) | \Phi(\theta_{\eta,\xi_0})(\beta)) \\
&= (\Phi(\theta_{\xi_0, \eta} \theta_{\xi,\xi_0})(\alpha) | \beta ) \\
&= (\xi|\eta) (\Phi(\theta_{\xi_0,\xi_0})(\alpha) | \beta )
= (\xi|\eta) (p(\alpha) | \beta ) \\
&= (\xi|\eta) (\alpha | \beta ).
\end{align*}
Thus $U$ is an isometry, and so extends to $H\otimes K_0$. Now compute
\begin{align*}
U^* \Phi(\theta_{\xi,\eta}) U(\xi_1\otimes\alpha)
&= U^* \Phi(\theta_{\xi,\eta}) \Phi(\theta_{\xi_1,\xi_0})(\alpha) \\
&= (\xi_1|\eta) U^*\Phi(\theta_{\xi,\xi_0})(\alpha) \\
&= (\xi_1|\eta) U^*U(\xi\otimes\alpha) \\
&= \theta_{\xi,\eta}(\xi_1) \otimes \alpha.
\end{align*}
So $U^*\Phi(\theta_{\xi,\eta})U = \theta_{\xi,\eta}\otimes 1$ and so by linearity and continuity, $U^*\Phi(\theta)U = \theta\otimes 1$ for each $\theta\in\mc K(H)$.

If we can show that $U$ has dense range, it must be onto (as it's an isometry), and so will be a unitary, and so $UU^*=1$ and so $\Phi(\theta) = U(\theta\otimes 1)U^*$ as required.

If $\xi_1$ is another vector, we see that
$$ \Phi(\theta_{\xi,\xi_1})(\alpha) = \Phi(\theta_{\xi,\xi_0})\Phi(\theta_{\xi_0,\xi_1})(\alpha) = U(\xi \otimes \beta), $$
say, where $\beta = \Phi(\theta_{\xi_0,\xi_1})(\alpha)$. Letting $\xi, \xi_1,\alpha$ vary, taking linear span, and using non-degeneracy, we see that $U$ does indeed have dense range.