Answer
The muon lepton number is conserved.
Work Step by Step
The proton $p$ has $L_\mu = 0$
The antiproton $\bar{p}$ has $L_\mu = 0$
The $\Lambda^0$ has $L_\mu = 0$
The $\Sigma^+$ has $L_\mu = 0$
The $e^-$ has $L_\mu = 0$
The left side of the proposed reaction has net $L_\mu = 0$ and the right side of the proposed reaction has net $L_\mu = 0$
Therefore, the muon lepton number is conserved.
