In the given reaction,
${\mathrm{MnO}}_2+4\mathrm{HCl}\to {\mathrm{MnCl}}_2+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{Cl}}_2$
$\mathrm{HCl}$ is getting oxidized to ${\mathrm{Cl}}_2$ and ${\mathrm{MnO}}_2$ is reduced to ${\mathrm{MnCl}}_2$. This can be explained as follows:Here,
$2\mathrm{e}+\stackrel{+4_2}{\mathrm{MnO}}\to {\stackrel{+2}{\mathrm{MnCl}}}_2$
$\mathrm{Mn}$ is going from +4 oxidation state to +2 oxidation thus clearly $\mathrm{Mn}$ is gaining electrons here ${\mathrm{MnO}}_2$ is reduced to ${\mathrm{MnCl}}_2$
$2\stackrel{-1}{\mathrm{HCl}}\to \stackrel{\mathrm{O}}{{\mathrm{Cl}}_2}+2{\mathrm{e}}^{-}$
Hence $\mathrm{Cl}$ of $\mathrm{HCl}$ is in -1 oxidation state and after losing its one electron it goes to 0 (elementary form) oxidation state, hence here $\mathrm{HCl}$ is oxidised to ${\mathrm{Cl}}_2$.