 Solution

Use $K^{-}=s\bar{u}$ and $p=uud$. In the first reaction a proton $u$ can join the incoming $\bar{u}$ to form the $u\bar{u}$ component of the neutral pion, leaving $uds$. In the second, a $u\bar{u}$ pair can annihilate into a gluon which creates $s\bar{s}$; the remaining $u$ joins $\bar{s}$ in the outgoing kaon, leaving $dss$. Thusbaryon number is one in both cases. Strong strangeness conservation provides a separate check: the first baryon has strangeness $-1$; the second has strangeness $-2$ because the outgoing $K^{+}$ carries $+1$. The Feynman diagrams show representative quark-line flavour rearrangement and pair-annihilation topologies. Bound hadrons require nonperturbative binding, so a single drawn gluon is not an exclusive cross-section calculation.