$4\mathrm{HCl}+\mathrm{A}\to {\mathrm{MnCl}}_2+{\mathrm{Cl}}_2\uparrow +2{\mathrm{H}}_2\mathrm{O}$. In this reaction the change in  oxidation state of metal present in A is

$4\mathrm{HCl}+\mathrm{A}\to {\mathrm{MnCl}}_2+{\mathrm{Cl}}_2\uparrow +2{\mathrm{H}}_2\mathrm{O}$, from the given reaction it is concluded that the product contains manganese as ${\mathrm{MnCl}}_2$ and oxygen as water molecule. So the reactant A might be metal manganese with oxygen - ${\mathrm{MnO}}_2$. Therefore the reaction is  

$4\mathrm{HCl}+{\mathrm{MnO}}_2\to {\mathrm{MnCl}}_2+{\mathrm{Cl}}_2\uparrow +2{\mathrm{H}}_2\mathrm{O}$

Here manganese metal gets reduced from $+4$ state in ${\mathrm{MnO}}_2$ to $+2$ in  the product ${\mathrm{MnCl}}_2$. 

Hence the correct option is (D).