To solve this problem, we need to identify which gas can readily decolorize acidified KMnO$_4$ solution.$\\$Acidified KMnO$_4$ is a strong oxidizing agent. It is commonly used to test for the presence of reducing agents.$\\$Let's analyze each option:$\\$• Option 1: SO$_2 \\$Sulfur dioxide (SO$_2$) is a well-known reducing agent. It can readily decolorize acidified KMnO$_4$ solution by reducing the $\\$purple permanganate ion (MnO$_4^-$) to the colorless manganese ion (Mn$^{2+}$).$\\$The reaction is as follows:$\\ 2$MnO$_4^- + 5$SO$_2 + 2$H$_2$O$\rightarrow 2$Mn$^{2+} + 5$SO$_4^{2-} + 4$H$^+ \\$• Option 2: NO$_2 \\$Nitrogen dioxide (NO$_2$) is not a strong reducing agent and does not readily decolorize acidified KMnO$_4$ solution.$\\$• Option 3: P$_2$O$_3 \\$Phosphorus trioxide (P$_2$O$_3$) is not typically involved in reactions with acidified KMnO$_4$ solution.$\\$• Option 4: CO$_2 \\$Carbon dioxide (CO$_2$) is not a reducing agent and does not decolorize acidified KMnO$_4$ solution.$\\$Based on the above analysis, the correct answer is Option 1: SO$_2$.$\\$Therefore, the gas that can readily decolorize acidified KMnO$_4$ solution is $$SO$_2$.