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This is an identity that comes up very often while I'm working with Boolean logic, and recently, it got me thinking. The method I use usually to prove it (at least) seems to rely on the Law of the Excluded Middle. I wonder if this identity holds in Intuitionistic logic as well. Thanks in advance for any help.

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No; it has the Law of the Excluded Middle as a special case.

Specifically, $q$ be $\top$ (that is, truth). We get $p\lor(\top\land\lnot p)\iff(p\lor\top)$, which simplifies to $p\lor\lnot p$, the Law of Excluded Middle (which we know doesn't hold in intuitionistic logic).

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    Thanks for that! I knew that the proof used the Law of Excluded Middle, but I never realized that the statement itself implied it. Thanks. – 2017-01-05
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No. The following Agda code shows that the $\implies$ direction is fine, but the other direction constructively implies double negation elimination. Basically, setting $Q$ to any provable proposition leads to $P\lor \lnot P$ pretty much immediately.

data ⊥ : Set where

data _∨_ (A B : Set) : Set where
    Left : A → A ∨ B
    Right : B → A ∨ B

¬_ : Set → Set
¬ A = A → ⊥

record _∧_ (A B : Set) : Set where
    constructor _,_
    field
        fst : A
        snd : B

l1 : {P Q : Set} → P ∨ (Q ∧ (¬ P)) → P ∨ Q
l1 (Left p) = Left p
l1 (Right (q , _)) = Right q

l2 : {R : Set} → ({P Q : Set} → P ∨ Q → P ∨ (Q ∧ (¬ P))) → ¬ (¬ R) → R
l2 {R} l nnr with l {R} {¬ (¬ R)} (Right nnr)
l2 {R} l nnr | Left r = r
l2 {R} l nnr | Right (_ , nr) with nnr nr
l2 {R} l nnr | Right (_ , nr) | ()