Let $A$ be a unital C*-algebra and $\phi: A\rightarrow M_{n}(\mathbb{C})$ be a completely positive map. If $P$ denotes the projection onto the kernel of $\phi(1_{A})$ and $P^{\perp}=1-P$ is the orthogonal complement.
My question is: can we conclude that $$\phi(a)=P^{\perp} \phi(a)= \phi(a)P^{\perp}$$ for all $a\in A$ ?
This works for any cp $\phi:A\to B(H)$. If $x\in \ker \phi(1_A)$, then for any $a\in A$, $$ \|\phi(a)x\|^2=\langle \phi(a)^*\phi(a)x,x\rangle\leq\|\phi(1_A)\|\,\langle\phi(a^*a)x,x\rangle\leq\|\phi(1_A)\|\,\langle\phi(\|a^*a\|\,1_A)x,x\rangle\\=\|\phi(1_A)\|\,\|a\|^2\,\langle\phi(1_A)x,x\rangle=0. $$ So $\phi(a)P=0$ for all $a\in A$. This implies that $\phi(a)P^\perp=\phi(a)$ for all $a\in A$. As this holds for all $a\in A$ and we can take adjoints, we also have $\phi(a)=P^\perp\phi(a)$.
This idea works even if you don't assume that the codomain of $\phi$ is $B(H)$. What I mean is that if $\phi(1_A)T=0$, then $\phi(a)T=0$ for all $a\in A$. Indeed, $$ (\phi(a)T)^*\phi(a)T=T^*\phi(a)^*\phi(a)T\leq \|\phi(1_A)\|\,T^*\phi(a^*a)T\leq \|\phi(1_A)\|\,\|a\|^2\,T^*\phi(1_A)T=0, $$ so $\phi(a)T=0$.