Answering to your edit: it is alternative a), your second code snippet is not a valid haskell value. The valid Haskell version would be:

    purelyFunctional :: IO Boolean
    purelyFunctional = do
        value1 <- notPurelyFunctional 5.0  
        value2 <- notPurelyFunctional 5.0  
        value1 == value2

Which, incidentally, is a pure value.