A $C^*$-algebra is postliminal if $\pi(A)\supset \mathcal K(H)$ for any irreducible representation $(\pi, H)$, where $\mathcal K$ denotes the compact operator algebra on $H$.
$B(l^2)$ is not postliminal.
Since $\pi(A)x$ is separable and invariant, $H$ is separable. Therefore $\pi$ is an homomorphism from $B(l^2)$ to some subalgebra of $B(l^2)$.
Assume $\pi(A)\not\supset \mathcal K$, then $\pi(A)\cap \mathcal K$ must be $0$ (because of irreducibility). However, let $p$ be a minimal projection, $\pi(p)$ is a projection. Assume the range of $\pi(p)$ contains at least $2$ nonzero diagonal vectors, say, $x,y$, again by irreducibility, there is some $\pi(a)\in \pi(A)$ such that $\pi(a)x=x$ and $\pi(a)y=0$. $pap=kp$ in $A$, however, $\pi(p)\pi(a)\pi(p)\neq kp$ since $\pi(p)\pi(a)\pi(p)x=x$ and $\pi(p)\pi(a)\pi(p)y=0$, which is a contradiction.
Where am I wrong?
You are assuming that $\pi(p)\ne0$. If you consider $\pi$ the quotient map onto the Calkin algebra (and you compose with an irrep of the Calkin algebra) you get an irreducible representation that is zero on the compacts of $\ell^2$; as the image does not contain minimal projections, it cannot contain the compacts of $H$.