Answer
$\text { With } $
$B(p)=+1, B(\bar{p})=-1, \quad B\left(\Lambda^{0}\right)=1, \quad B\left(\Sigma^{+}\right)=+1, \text { and } B\left(e^{-}\right)=0,$
we have
$$1+(-1) \neq 1+1+0 . \text { Thus, the process does not conserve baryon number. } $$
Work Step by Step
$\text { With } $
$B(p)=+1, B(\bar{p})=-1, \quad B\left(\Lambda^{0}\right)=1, \quad B\left(\Sigma^{+}\right)=+1, \text { and } B\left(e^{-}\right)=0,$
we have
$$1+(-1) \neq 1+1+0 . \text { Thus, the process does not conserve baryon number. } $$
