I am asking why it is not possible for two pions to have CP=-1? Griffiths says that part as a footnote to the mentioned section. I know I am missing something important, and that's why I am asking this. So are you saying two pions, when we consider their relative orbital angular momenta can not be antisymmetric? But for three pions it can be symmetric or antisymmetric?
What footnote? Be more specific; I suspect an impossible rabbit hole. Two pions cannot have CP=-1. But as @anna v reminds you, both $𝐾_{1,2}$ have spin 0 and decay to an s-state, (L=0); so in symmetric combinations of pions.
Two such pions have P=+, and three have P= - . In both cases, you have C= + . So, CP=+ for 2 pions (the daughters of $K_1$, the short-lived θ); and - for 3 pions, (daughters of $K_2$, the long-lived τ). Griffiths hammers this in in (4.70), and the sentence right above it, "$K_1$ decays into 2π (never three); $K_2$ decays into 3π (never two)."
(But since you appear obsessed with this irrelevancy, other particles decay to 2 pions, as well, at higher spins, like p-wave (L=1): the ρ meson. It has P=-, C=-, hence CP=+ again! The parity switch and the charge interchange, e.g., for $\pi^+\pi^-$, cancel each other. Go down the PDG listings and look for decays violating this systematic rule—in vain.)
Post OP-edit-footnote commentary
With your edit, that you should have included right away in your question, your puzzlement appears less preposterous.
Above, I'm quoting from the 1st edition, while you are quoting (and being needlessly confused by) the second, revised edition (2008). A truly unfortunate hedging footnote, as it tripped you into the rabbit hole. Still, it states expressly that the "same logic" cannot apply, virtually by conservation of angular momentum, as already described above: there is no way to add orbital L to two spin 0 particles to get a spin 0 particle with negative CP: the two pions must be in an s-wave.
Nevertheless, for academic completeness, I can expand to shed light on the logic of the ("actually...") defensive footnote, addressed to fussbudget students, and trusting (?) it won't confuse them... Indeed, there is a very rare $\pi^+\pi^-\pi^0$ decay mode of $K_1\sim K_S$, with BR a million times smaller than the 2π mode, with mixed symmetry, so not a necessarily fully symmetric one like $3\pi^0$ which must violate CP! (We are assuming CP conservation throughout in this setup.) It is the smallest $K_s$ decay mode inferred and reported (CPLEAR at CERN), Sec 2.2.3. Note isospin is not conserved in this weak decay: the relevant part of the weak hamiltonian yields $\Delta I =3/2$ to reach an I=1 final state. (The mixed space symmetry is matched by the mixed isospin symmetry to amount to an overall Bose symmetry, as they should: Charged pion mutual interchange antisymmetry, charged-neutral pion interchange symmetry, $(\pi^+ \pi^{-} -\pi^-\pi^+)\pi^{0} -\pi^0(\pi^+\pi^{-} -\pi^-\pi^+)$.)
As we saw above, a $\rho^0 ~~(J^{PC}=1^{--})$ is thinkable as a p-wave (L=1) combination of $\pi^+ \pi^-$; presumably you have learned that it does not decay to $\pi^0\pi^0$ and why (antisymmetry). Think of combining this $\rho^0$ a with a $\pi^0~~ (J^{PC}=0^{-+})$ also in a p-wave, so with L=1, to a spin 0 state flipping P and resulting in a $J^{PC}=0^{--}$ extremely exotic state, so with PC=+, the conventional "nono" for three pions.
(But I am a little out of my depth here: the $a_0(980)~~ (J^{PC}=0^{++})$ also has CP=+ and, being G-parity odd, can decay to 3π, although I know little about its chiral lagrangian couplings. In any case, it would only figure in a footnote of a footnote of your footnote... If you are really, really curious, try the appendix of this unpublished preprint.)
