Let $A$ and $B$ be $C^*$-algebras such that $A$ is unital, and let $\phi: A\rightarrow B$ be a c.p.c. order zero map. Then $\phi$ is a $*$-homomorphism if and only if $\phi(1_A)$ is a projection.
Since $\phi$ is a c.p.c. order zero map, we know that $\phi(ab)\phi(1_A)=\phi(a)\phi(b)$ for all $a,b \in A$. If $\phi$ is a $*$-homomorphism, it is obvious that $\phi(1_A)$ is a projection. How to prove the the other direction?
From the identity $\phi(ab)\phi(1_A)=\phi(a)\phi(b)$, it suffices to show that $\phi(a)\phi(1_A)=\phi(a)$ for all $a$. Since positive elements span $A$, it suffices to show this for $a\geq 0$.
First, note that $\phi(1_A)$ commutes with $\phi(A)$. Using the $C^*$-identity, the fact that $\phi(1_A)$ is a projection, and the identity $\phi(ab)\phi(1_A)=\phi(a)\phi(b)$, we obtain \begin{align*} \|\phi(a)\phi(1_A)-\phi(a)\|^2&=\|(\phi(a)\phi(1_A)-\phi(a))(\phi(a)\phi(1_A)-\phi(a))\|\\ &=\|\phi(a)\phi(1_A)\phi(a)\phi(1_A)-\phi(a)\phi(1_A)\phi(a)-\phi(a)^2\phi(1_A)+\phi(a)^2\|\\ &=\|\phi(a)^2-\phi(a)^2\phi(1_A)\|\\ &=\|\phi(a^2)\phi(1_A)-\phi(a^2)\phi(1_A)\|\\ &=0. \end{align*}