Answer
since
$$B\left(\Xi^{-}\right)=+1,$$ $$B\left(\pi^{-}\right)=0,$$ $$B(\mathrm{n})=+1,$$ $$B\left(\mathrm{K}^{-}\right)=0,$$ and $$B(\mathrm{p})=+1,$$
we have
$$+1 \neq 0+1+0+1=2 .$$
$$ \text { Thus, process does not conserve (baryon_number). } $$
Work Step by Step
since
$$B\left(\Xi^{-}\right)=+1,$$ $$B\left(\pi^{-}\right)=0,$$ $$B(\mathrm{n})=+1,$$ $$B\left(\mathrm{K}^{-}\right)=0,$$ and $$B(\mathrm{p})=+1,$$
we have
$$+1 \neq 0+1+0+1=2 .$$
$$ \text { Thus, process does not conserve (baryon_number). } $$
