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Despite the comments to the question (including mine), this is a bit easier than it seems at first sight. We can show that $L_{\max}=2$ or $3$. Almost certainly we have $L_\max=3$. However, determining which of these is actually the case seems to be beyond current technology, according to this MO answer "Are all primes in a PAP-3?".

Showing that, $L_{\max} < 4$ is easy. That is, not every odd prime is contained in an arithmetic progression of primes of length 4. More specifically, 3 is not contained in an arithmetic progression of length 4. Suppose that $\lbrace x, x+d, x+2d, x+3d \rbrace$ was such a progression for $d > 0$. Then $x\not=2$, otherwise we would have $x=2,d=1$, but $x+2d=4$ is not prime. So, $x=3$. But, then, $x+3d=3(1+d)$ is not prime.

Edit: Looking at $\tilde L_\max \equiv \max_A\liminf_i \vert A_i\vert$ might be more interesting. I expect that this is infinite but, again, showing that $\tilde L_\max > 2$ appears to be beyond current means.
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Despite the comments to the question (including mine), this is a bit easier than it seems at first sight. We can show that $L_{\max}=2$ or $3$. Almost certainly we have $L_\max=3$. However, determining which of these is actually the case seems to be beyond current technology, according to this MO answer "Are all primes in a PAP-3?".

Showing that, $L_{\max} < 4$ is easy. That is, not every odd prime is contained in an arithmetic progression of primes of length 4. More specifically, 3 is not contained in an arithmetic progression of length 4. Suppose that $\lbrace x, x+d, x+2d, x+3d \rbrace$ was such a progression for $d > 0$. Then $x\not=2$, otherwise we would have $x=2,d=1$, but $x+2d=4$ is not prime. So, $x=3$. But, then, $x+3d=3(1+d)$ is not prime.
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Despite the comments to the question (including mine), this is a bit easier than it seems at first sight. We can show that $L_{\max}=2$ or $3$. Almost certainly we have $L_\max=3$. However, determining which of these is actually the case seems to be beyond current technology, according to this MO answer "Are all primes in a PAP-3?".

Showing that, $L_{\max} < 4$ is easy. That is, not every prime is contained in an arithmetic progression of primes of length 4. More specifically, 3 is not contained in an arithmetic progression of length 4. Suppose that $\lbrace x, x+d, x+2d, x+3d \rbrace$ was such a progression for $d > 0$. Then $x\not=2$, otherwise we would have $x=2,d=1$, but $x+2d=4$ is not prime. So, $x=3$. But, then, $x+3d=3(1+d)$ is not prime.