I would add to Taka's answer, since the Choi-Effros paper is not widely available online, that the only nontrivial part of the proof is to show the "C$^*$-identity" $\|x\|^2=\|x^*x\|$.
Indeed, the algebra $(P(A),+,\circ)$ is a $*$-algebra, normed with the norm inherited from $A$. For the C$^*$-identity, since $P$ is contractive we have
\[ \|x^*\circ x\|=\|P(x^*x)\|\leq\|x^*x\|=\|x\|^2. \]
On the other hand, since $P$ is cp and contractive, it satisfies the Schwarz inequality $P(x)^*P(x)\leq P(x^*x)$, and so, for any $x\in P(A)$,
\[ \|x^*\circ x\|=\|P(x^*x)\|\geq \|P(x)^*P(x)\|=\|x^*x\|=\|x\|^2. \]