 Solution

For both modes, a leading weak quark transition in $K^{+}=u\bar{s}$ is $\bar{s}\to \bar{u}+W^{+}$, followed by $W^{+}\to u\bar{d}$, with the original $u$ acting as a spectator quark. This produces the four quarks needed for ${\pi }^{+}{\pi }^0$, using the $u\bar{u}$ component of ${\pi }^0$. Producing three pions additionally requires a $d\bar{d}$ pair from strong hadronization. A representative quark-line diagram draws it as gluon emission and splitting.

The three-pion assignment is $u\bar{d}$, $u\bar{d}$, and $d\bar{u}$, matching ${\pi }^{+}{\pi }^{+}{\pi }^{-}$. The extra soft pair creation is a schematic representation of hadronization, not a controlled perturbative prediction of the relative decay rates. Both processes are first order in the effective weak interaction.

There is an explicit error or hypothetical assumption to address: the original PDF assigns $0^{+}$ to all these mesons. Taken literally, the two-pion final state has $L=0$, because all particles have zero spin and the initial $J$ is zero. Its parity is $(+1{)}^2(-1{)}^0=+1$. For three spinless particles, use two Jacobi orbital angular momentum quantum numbers $l,L$. Coupling to total zero requires $l=L$, so the orbital parity $(-1{)}^{l+L}$ is $+1$ and the three-particle parity is also $+1$. Thus under the printed $0^{+}$ assignment, both channels conserve parity.

The named physical pions and kaons are instead pseudoscalar mesons with $J^P=0^{-}$. With those physical assignments,ThereforeThese two opposite-parity final states from the same physical kaon are a classic illustration of parity violation in a weak decay. The physical conclusion requires correcting the printed parity assignment explicitly.