Visual Studio loading the right (x86 or x64) dll

This slightly simpler answer than Preet Sangha's will not generate a warning when the project is loaded and only the conditionally accepted dll will appear in the Solution Explorer. So, overall, the appearance is cleaner, although more subtle. (This was tested in Visual Studio 2010.)

<Reference Include="p4dn" Condition="$(Platform) == 'x86'">
  <HintPath>..\..\ThirdParty\P4.Net\clr4\x86\p4dn.dll</HintPath>
</Reference>
<Reference Include="p4dn" Condition="$(Platform) == 'x64'">
  <HintPath>..\..\ThirdParty\P4.Net\clr4\x64\p4dn.dll</HintPath>
</Reference>

in your project file in reference use an MSBUILD conditional

<Reference 
       Include="SomeAssembly86, Version=0.85.5.452, Culture=neutral, PublicKeyToken=41b332442f1101cc, processorArchitecture=MSIL"  
         Condition=" '$(Platform)' == 'AnyCPU' ">
      <SpecificVersion>False</SpecificVersion>
      <HintPath>..\..\Dependencies\SomeAssembly.dll</HintPath>
      <Private>False</Private>
    </Reference>
    <Reference 
         Include="SomeOtherAssembly, Version=0.85.5.999, Culture=neutral, PublicKeyToken=41b332442f1101cc, processorArchitecture=MSIL" 
         Condition=" '$(Platform)' == 'x64' ">
      <SpecificVersion>False</SpecificVersion>
      <HintPath>..\..\Dependencies\SomeOtherAssembly.dll</HintPath>
      <Private>False</Private>
    </Reference>