I don't have access to Pisier's memoir, but I think I can provide a rough sketch of the proof.
First step consists of showing that the assumptions imply that there exists a completely bounded projection $P:B(H) \to A^{\ast\ast}$. Then, and this is the hard part, one has to prove that it implies that $A^{\ast\ast}$ is injective, hence $A$ is nuclear. In fact, the assumptions imply that $A$ is exact and ijectivity of $A^{\ast\ast}$ immediately implies that $A$ has the WEP (weak expectation property), and it is not hard to show that combination of WEP and exactness (understood as existence of approximate factorisation of the embedding $A\subset B(H)$ through matix algebras) gives nuclearity.
In order to do that, we will work with maps $\phi_k^{\ast\ast}: A^{\ast\ast}\to M_{d_k}$ and $\psi_k: M_{d_k} \to A^{\ast\ast}$, whose composition converges to identity in point-ultraweak topology; it is not entirely clear to me that this is actually true. However, there is a way to produce a sequence that will tend to identity in point-ultraweak topology, using local reflexivity. Indeed, the assumptions say that $A$ has a strong form of completely bounded approximation property, therefore it is exact, hence locally reflexive. In particular, it means that whenever $X \subset A^{\ast\ast}$ is a finite dimensional subspace and $(\omega_1, \dots, \omega_n)\subset A^{\ast}$, then for any $\varepsilon > 0$ there is a completely contractive map $P: X \to A$ such that $|\omega_i(x-Px)| \leqslant \varepsilon \|x\|$. Suppose that we have $A\subset A^{\ast\ast} \subset B(H)$. If we compose such a $P$ with $T_k:= \psi_k \circ \phi_k$, then by by Arveson-Wittstock theorem these maps can be extended to maps from $B(H)$ to $A$, since $phi_k\circ P$ has values in $M_{d_k}$. These extensions may be viewied as completely bounded maps from $B(H)$ to $A^{\ast\ast}$, and we have $CB(B(H), (A^{\ast})^{\ast})\simeq B(H)\widehat{\otimes} A^{\ast}$, where $\widehat{\otimes}$ is the projective tensor product (of operator spaces). Therefore any bounded net would have a weak* cluster point. If we are careful enough, we may construct a net whose cluster point must be the identity on $A^{\ast\ast}$.
I will first describe the averaging approach of Christensen and Sinclair (see On von Neumann algebras which are complemented subspaces of $B(H)$, J. Funct. Anal. 122 (1994), no. 1, 91--102). The key is the following: suppose that $P:B(H) \to M$ is a completely bounded projection and $(u_i)$ is a sequence of elements in $M$ such that $\sum u_i^{\ast} u_i = \mathrm{Id}$ strongly. Then the ultraweakly closed convex hull of the maps of the form $x\mapsto \sum P(xu_{i}^{\ast}) u_i$ (indexed by such families of $u_i$'s) contains a completely bounded projection $Q$ which is a right $M$-module map. Doing it from the other side, we obtain an $M$-bimodule projection; this is enough by the results presented by Bunce and Paschke in Quasi expectations and amenable von Neumann algebras, Proc. Amer. Math. Soc. 71 (1978), 232--236. This is actually not very hard. If we assume that $M$ is a finite von Neumann algebra with trace $\tau$ then let $\varphi$ be a functional on $B(H)$ defined by $\varphi(x) :=\tau(Q(x))$, where $Q$ is a bimodular projection. Then $\varphi$ is unitarily invariant. We may assume that $\varphi$ is self-adjoint and then perform the Hahn-Jordan decompostion into positive and neqative parts, and uniqueness of this decomposition implies that both parts are unitarily invariant. If we take the positive part then we obtain a hypertrace on $B(H)$, which is sufficient for proving that $M$ is injective. The general case is handled by standard techniques, the semifinite case by approximation, and the type III case using the crossed product decompostion.
Pisier's (and Haagerup's?) approach is completely different, and he actually requires slightly less than complete boundedness, he requires that the projections works nicely with row and column spaces (see Projections from a von Neumann algebra onto a subalgebra, Bull. Soc. Math. France 123 (1995), 139--153). The main result is: let $M$ be a von Neumann algebra, and let $X_0$ ($X_1$) be $M^n$ equipped with the norm $\|(x_i)\|_{0}:=\|\sum x_i^{\ast}x_i\|_M^{\frac{1}{2}}$ ($\|(x_i)\|_{1}:=\|\sum x_i x_i^{\ast}\|_M^{\frac{1}{2}}$). The interpolation result is that $\|(x_i)\|_{\frac{1}{2}} = \|\sum \overline{x_i}\otimes x_i\|^{\frac{1}{2}}_{\overline{M}\otimes_{\mathrm{max}} M}$. In particular, even though general completely bounded maps do not interact nicely with the maximal tensor product, they do so if we restrict to ''positive definite'' tensors.
Suppose now that $N\subset M$ is a pair of von Neumann algebras such that there exists a completely bounded projection $P:M \to N$. By the interpolation result we have $\|\sum \overline{x_i}\otimes x_i\|_{\overline{M}\otimes_{\mathrm{max}} M} \leqslant \|P\|_{cb}^2 \|\sum \overline{x_i} \otimes x_i\|_{\overline{N} \otimes_{\mathrm{max}} N}$, for any tuple $(x_i)\subset N$. Note that if we denote $t=\sum \overline{x_i}\otimes x_i$, then $(t^{\ast}t)^m$ is a positive definite tensor as well for any $m$, and by $C^{\ast}$-identity we get the above inequality with constant $\|P\|_{cb}^{\frac{1}{m}}$ for any $m$, i.e. with constant $1$; it means that the inclusion $N \subset M$ is $\mathrm{max}^{+}$-injective. Then one can use the self-polar forms to show that in such case there exists a norm one projection from $M$ onto $N$. It would take some time to discuss it, but if anyone is interested I can elaborate on that.
The second approach seems way harder, but it gives more, e.g. Haagerup's characterisation of WEP algebras: a $C^{\ast}$-algebra $A$ is WEP iff the max and min norms coincide on positive definite tensors in $\overline{A}\otimes A$; this shows in particular that WEP depends only on the operator space structure of $A$.
Maybe a comment is in order; it seems to me that all of these results when something a priori completely bounded can be replaced by something completely contractive (or even completely positive), really come from interpreting various properties in terms of tensor products and then using the rigidity of the norm structure of a $C^{\ast}$-algebra, i.e. whenever we have equivalent norms they have to be equal.
I hope that this answer is helpful to someone; writing it certainly clarified matters a lot for me.
