Definition 2.1.1 Let $A, B$ be the C*-algebra, a map $\theta: A\rightarrow B$ is called nuclear if there exist contractive completely positive maps $\phi_{n}: A\rightarrow M_{k(n)}(\mathbb{C})$ and $\psi_{n}:M_{k(n)}(\mathbb{C}) \rightarrow B$ such that $$||\psi_{n}\circ\phi_{n}(a)-\theta(a)||\rightarrow0$$ for all $a\in A$.
Exercise 2.1.2 Prove that $\theta: A\rightarrow B$ is nuclear if and only if there exist contractive completely positive maps $\phi_{n}:A\rightarrow C_{n}$ and $\psi_{n}:C_{n} \rightarrow B$, where the $C_{n}$ are finite-dimensional C*-algebras, such that $\psi_{n} \circ\phi_{n}\rightarrow\theta$ in the point-norm topology. (Hint: Find integers $k(n)$ such that there is a unital embedding $C_{n}\subset M_{k(n)}(\mathbb{C})$ and construct a conditional expectation $M_{k(n)}(\mathbb{C}) \rightarrow C_{n}$.)
Could someone show me more details of this question?
One implication is obvious, since you can take $C_n=M_{k(n)}(\mathbb C)$.
So assume you have an approximate factorization of $\theta$ by $A\xrightarrow{\phi_n}C_n\xrightarrow{\psi_n}B$. Following the hint, choose $k(n)$ such that, for each $n$, $C_n\subset M_{k(n)}(\mathbb C)$.
Fix $n$. Being a finite-dimensional C$^*$-algebra, each $C_n$ is a sum of full-matrix blocks. This means that there exist pairwise orthogonal projections $P_1,\ldots,P_s$ in $M_{k(n)}(\mathbb C)$ such that $$ C_n=\sum_{j=1}^s P_jM_{k(n)}(\mathbb C)P_j. $$ So the map $\rho_n:X\mapsto \sum_j P_jXP_j$ is a completely positive map $M_{k(n)}(\mathbb C)\to C_n$ that is the identity when restricted to $C_n$ (i.e., a conditional expectation).
Now define $\psi'_n=\psi_n\circ\rho_n$. This is completely positive for all $n$, and now we have $A\xrightarrow{\phi_n} M_{k(n)}(\mathbb C)\xrightarrow{\psi'_n} B$.
For any $a\in A$, $\psi'_n(\phi_n(a))=\psi_n(\rho_n(\phi_n(a)))=\psi_n(\phi_n(a))$. So $\theta(a)=\lim_n\psi'_n\circ\phi_n(a)$.