Let $M$ be a model category, and suppose the projective model structure on $M^{\mathbb N_{\ge 0}}$. Then
The diagram $$A_0 \rightarrowtail A_1 \rightarrowtail \cdots$$ with cofibrations between cofibrant objects is projectively cofibrant.
I do not see how one this satisfies the LLP with respect to trivial fibrations in this model strucutre.
You can lift level wise but they need not commute. Let me show my struggle.
Let us suppose we have a diagram $$ A_\bullet \rightarrow C_\bullet$$ $$B_\bullet \rightarrow C_\bullet$$ and we wish to construct a lift $A_\bullet \rightarrow B_\bullet$. We define our first map $A_0 \rightarrow B_0$. Now to construct $A_1 \rightarrow B_1$, we consider the lift of $A_0 \rightarrow A_1$ against $B_1 \rightarrow C_1$. The top morphism being the composition $A_0 \rightarrow B_0 \rightarrow B_1$.
What confuses me is that I only used $A_0$ cofibrant.
Working in our model category $M$, let $p_\bullet:B\bullet\stackrel\simeq\twoheadrightarrow C_\bullet$ by a natural map which is a pointwise weak equivalence and a pointwise fibration (i.e. a projecive acyclic fibration). Assume that $A_\bullet$ is a $\mathbb{N}_{\geq0}$-indexed diagram of cofibrations $a_i$ between cofibrant objects $A_i$ and that $f_\bullet:A_\bullet\rightarrow C_\bullet$ is a natural map. In the sequel I'll write $a_i:A_i\rightarrow A_{i+1}$ for the component maps of $A_\bullet$. Similarly I'll write $b_i$ and $c_i$ for the component maps of $B_\bullet$ and $C_\bullet$.
To begin, since $A_0$ is cofibrant, there exists a lift $\hat f_0:A_0\rightarrow B_0$ satisfying $p_0\hat f_0=f_0$. Now consider the diagram $\require{AMScd}$ \begin{CD} A_0@>b_0\hat f_0>> B_1\\ @Va_0V V @VV p_1 V\\ A_1 @>f_1>> C_1. \end{CD} Since $p_1b_0\hat f_0=c_0p_0\hat f=c_0f_0=f_1a_0$ the diagram commutes. The left-hand map $a_0$ is a cofibration and $p_1$ on the right is an acyclic fibration. Hence there exists a lift $\hat f_1:A_1\rightarrow B_1$ satisfying $\hat f_1 a_0=b_0\hat f_0$ and $p_1\hat f_1=f_1$.
Assume by induction that we have constructed $\hat f_n:A_n\rightarrow B_n$ satisfying $p_n\hat f_n=f_n$ and $\hat f_n a_{n-1}=b_{n-1}\hat f_{n-1}$. Then simply reindexing the previous diagram we present another solvable lifting problem for a map $\hat f_{n+1}:A_{n+1}\rightarrow B_{n+1}$, and our inductive step follows.