It is known that for every $k\in\mathbb{N}$ there is $N_k\in\mathbb{N}$ such that every $N_k$-dimensional subspace of $\ell_p$, $1<p<\infty$, contains uniformly complemented copies of $\ell_2^k$ (see Theorem 4.10 in this paper). It is also known that the Lorentz sequence space $d(a,p)$ contains a complemented copy of $\ell_p$ formed from a block basis of the canonical basis $(x_n)_{n=1}^\infty$ of $d(a,p)$. (Definitions and a lot of nice results on Lorentz sequence spaces can be found in this paper.)
Hence, we could find some uniform constant $C\in[1,\infty)$ such that for each $k\in\mathbb{N}$ there is $N_k\in\mathbb{N}$ and a norm-$C$ projection $P_k:[x_n]_{n=1}^{N_k}\to[x_n]_{n=1}^{N_k}$ onto a $C$-isomorphic copy of $\ell_2^k$.
Now, let's string together these uniformly complemented copies of $\ell_2^k$ in the following way. We "shift" (and relabel) $P_k$ so that it acts on $[x_n]_{n=M_k+1}^{M_k+N_k}$, where $M_k=\sum_{i=1}^{k-1}N_i$. Now, let $(v_n^{(k)})_{n=1}^k$ denote the copy of the canonical basis of $\ell_2^k$ sitting in $[x_n]_{n=M_k+1}^{M_k+N_k}$. Without loss of generality, assume it is normalized. Note that $\{(v_n^{(k)})_{n=1}^k:k\in\mathbb{N}\}$, now forms a normalized block basic sequence in $d(a,p)$, with each consecutive $k$-long stretch uniformly isomorphic to the $\ell_2^k$ basis, and uniformly complemented in $d(a,p)$.
My question is: could the block basic sequence $\{(v_n^{(k)})_{n=1}^k:k\in\mathbb{N}\}$ span a complemented subspace of $d(a,p)$?
It seems plausible since it is formed from uniformly complemented stretches of vectors. But, the basis of $d(a,p)$ is not always as well-behaved as we might like, so maybe this is not true.
Thank you!