This is not really an answer, but an idea. I am not sure if it works, plus you need an extra assumption on the completely positive map. **Definition:** We call a completely positive map $\psi \colon A \to B$ **of order zero** if it is orthogonality preserving, i.e. if $ab = 0$ implies that $\psi(a)\,\psi(b) = 0$ for any $a,b \in A$. Let $A$ and $B$ be unital $C^*$-algebras and let $\psi \colon A \to B$ be a completely positive *order zero* map. There is a powerful structure theorem for these maps, which can be found in Section 3.3 of the paper "[Completely positive maps of order zero][1]" by Wilhelm Winter and Joachim Zacharias: Let $C = C^*(\psi(A))$ be the $C^*$-algebra generated by the image of $\psi$, then $h = \psi(1_A) \in C$ commutes with $C$ and there is a $*$-homomorphism $\pi_{\psi} \colon A \to M(C) \cap \{h\}'$ such that $$ h^{1/2}\,\pi_{\psi}(a)\,h^{1/2} = h\,\pi_{\psi}(a) = \pi_{\psi}(a)h = \psi(a) $$ for all $a \in A$. This says that completely positive order zero maps can always be written as compressions of a $*$-homomorphism into a bigger algebra by a positive element. If you now start with a projection $p \in A$ (or in a matrix algebra over $A$), then you can consider the sequence $$ q_n = \left(h +\frac{1}{n}\right)^{-1}\psi(p) . $$ This should get arbitrarily close to a projection in $B$ and hence can be cut down to an honest projection for $n$ big enough. [1]: http://wwwmath.uni-muenster.de/mjm/vol_2/mjm_vol_2_14.pdf