$Ph-C\equiv C-Me\xrightarrow{H_2,\ Pd/C\ (Lindlar's\ catalyst)}$ cis-$PhCH=CHMe$ (K); $\xrightarrow{Na/liq.\ NH_3}$ trans-$PhCH=CHMe$ (L). So K and L are geometrical isomers. K + HBr gives $Ph-CH(Br)-CH_2-CH_3$ (M); L + HBr/benzoyl peroxide gives $Ph-CH_2-CH(Br)-CH_3$ (N), which are position isomers, not stereoisomers.