In the given reaction, $\mathrm{MnO}_2+4\mathrm{HCl}\rightarrow\mathrm{MnCl}_2+2\mathrm{H}_2\mathrm{O}+\mathrm{Cl}_2$$\mathrm{HCl}$ is getting oxidized to $\mathrm{Cl}_2$ and $\mathrm{MnO}_2$ is reduced to $\mathrm{MnCl}_2$. This can be explained as follows:Here, $2\mathrm{e}+\overset{+4_2}{\mathrm{MnO}\rightarrow\overset{+2}{\mathrm{MnCl}_2}}$ $\mathrm{Mn}$ is going from +4 oxidation state to +2 oxidation thus clearly $\mathrm{Mn}$ is gaining electrons here $\mathrm{MnO}_2$ is reduced to $\mathrm{MnCl}_2$$2\overset{-1}{\mathrm{HCl}}\rightarrow\overset{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 $\mathrm{Cl}_2$.