I do understand this is an old question but, considering that:
1. you might still be interested in a simpler (and surely less elegant) proof;
2. I had the same problem, so this might be helpful for others in the future
I'll tell you how I solved this problem

Take your *-homomorphism $\lambda:M_n\to B$, where $B$ is any other C*-algebra. Set
$$F_{ij}:=\lambda(E_{ij}),\qquad\forall i,j=1,\ldots,n,$$
where the $E_{ij}$ is the canonical basis of the underlying vector space of $M_n$ (then the $F_{ij}$ satisfy the very same algebra, i.e. $F_{ij}F_{mn}=\delta_{jm}F_{in}$), and uppose that the kernel of $\lambda$ is non-trivial. Then there exists $a\in M_n\smallsetminus \{0\}$ s.t. $\lambda(a)=0$. This is a constraint between all the $F_{ij}$, namely there are coefficients $\alpha_{ij}$ such that
$$\sum \alpha_{ij}F_{ij}=0.$$
Note that if just one of the $F_{ij}$ for some $(i,j)$ is 0, then $\lambda$ is 0, because $F_{kk}$ are all Murray-von Neumann equivalent projections, and the "off-diagonal" elements $F_{ij}$, $i\neq j$ are partial isometries linking them. Therefore sandwich the above linear combination between $F_{kk}$ and $F_{mm}$ to obtain
$$0=\sum_{ij}\alpha_{ij}F_{kk}F_{ij}F_{mm} = \alpha_{km}F_{km},$$
which implies
$$\alpha_{km}=0$$
for arbitrary $(k,m)$. Hence all the $F_{ij}$ are linearly independent, meaning that, as a vector space, $B$ must have enough space to accommodate at least a copy of $M_n$. But taking into account once again that all the projections $F_{kk}$ are Murray-von Neumann equivalent, we can conclude that there must exists a subalgebra $C$ of $B$ such that $M_n\otimes C\subset B$ and a projection $P\in C$ such that, up to unitary equivalence
$$\lambda(E_{ij})=E_{ij}\otimes P\in M_n\otimes C.$$
If the rank of $P$ is $k$, then $B$ should be large enough to accommodate $k$ copies of $M_n$, i.e. the underlying vector space of B must have dimension greater than $kn$ (if it is not exactly $kn$ the you'll have some 0-padding).

Hope this helped like it did for me!