Let $A$ be C*-algebra then we know that $M_n(A)$ is also a C*-algebra.
Let $\rho:M_n(A)\rightarrow B(K)$ be a *-representation of $M_n(A)$ on some Hilbert space $K$. Then there exists a *-representation $\pi:A\rightarrow B(H)$ of $A$ on a Hilbert space $H$ such that
1) $H^{(n)}=K$, that is, the $H\oplus ... \oplus H$ ($n$-times) is $K$, and
2) $\pi^{(n)}=\rho$, that is, the map $\pi^{(n)}:M_n(A)\rightarrow B(H^{(n)})$ defined by $$\pi^{(n)}([a_{ij}])=[\pi(a_{ij})],$$ equals $\rho$.
I am unable to figure out a way to construct $H$ from $K$ and define the map $\pi$. Any hints are welcome.
Assume for simplicity that $A$ is unital. If you had that $K=H^{(n)}$, then you know that the projections onto each copy of the direct sum would be $E_{jj}$, where $E_{kj}$ denote the canonical matrix units in $M_n(A)$ (this where "unital" is useful, but it can be dispensed with).
Also, your last formula suggests what you need to do: you can define $H=E_{11}K$ and $\pi$ by $$ \pi(a)=\rho(a\otimes E_{11})|_{E_{11}K}. $$ There are some identifications to be checked: for an easy one, the subspaces $E_{jj}K$ and $E_{kk}K$ are isomorphic because $E_{kj}$ implements the unitary.