Lindlar's catalyst cannot be used to carry out the following process

a) ${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3\stackrel{{\mathrm{H}}_2}{⟶}$ trans Butene-2

b) ${\mathrm{CH}}_3-\mathrm{C}=\mathrm{C}-{\mathrm{CH}}_3\stackrel{{\mathrm{H}}_2}{⟶}$ Butane

c) ${\mathrm{C}}_2{\mathrm{H}}_2\stackrel{2{\mathrm{H}}_2}{⟶}{\mathrm{C}}_2{\mathrm{H}}_6$

d) ${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3\stackrel{{\mathrm{H}}_2}{⟶}$ cis Butene-2

The reactions are as follows:

a) ${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3\stackrel{\mathrm{Na} \mathrm{in} \mathrm{liq}. {\mathrm{NH}}_3}{⟶}$ trans Butene-2

b) ${\mathrm{CH}}_3-\mathrm{C}=\mathrm{C}-{\mathrm{CH}}_3\stackrel{\mathrm{Pt}}{⟶}$ Butane

c) ${\mathrm{C}}_2{\mathrm{H}}_2\stackrel{\mathrm{Pt} \mathrm{or} \mathrm{Ni}}{⟶}{\mathrm{C}}_2{\mathrm{H}}_6$

d) ${CH}_3-C\equiv C-{CH}_3\stackrel{Pd-BaS{O}_4}{⟶}$ cis Butene-2

Hence the correct option is (A).