$\ce{CH CH CH OH ->[Conc \, H_2SO_4][160^{\circ} - 180^{\circ} C] CH_3CH = CH_2}$
$\ce{ ->[Br_2] ->[alc. KOH] $\underset{\text{ (Y) }}{\ce{ CH_3CHBr}}$ - CH_2Br }$
$\ce{[$\underset{\text{ (A) (B) }}{\ce{CH3C(Br) = CH2 }}$ + CH3CH = CHBr ] }$
$\ce{ ->[NaNH_2][-HBr] CH_3 C \equiv CH }$
Alcoholic $KOH$ brings about dehydrobromination of $Y$ and give a mixture of vinyl bromide ($A$ and $B$) while $NaNH_2$ being a strong base than $alc.KOH$ readily brings about dehydrobromination of less reactive vinyl bromide to give propyne $\ce{CH3 C \equiv CH}$ i.e. $(Z)$.
