Since "implication" in Boolean Algebra means "the complement of the union" (or C==AN if you know Polish notation) you want to show
([(P∧Q)≡R]'V[(P→R)∨(Q→R)]) as an identity for any Boolean Algebra. Boolean Algebra has a very powerful metatheorem which states that if an identity holds for the two-element Boolean Algebra ({0, 1}, V, ∧, ') it will hold as an identity for any Boolean Algebra. So, if we can show that (([(P∧Q)≡R]'V[(P→R)∨(Q→R)])≡1) for {0, 1}, then we'll have shown ([(P∧Q)≡R]'V[(P→R)∨(Q→R)]) as a theorem for any Boolean Algebra.
Case 1: Suppose P=0. Then we have ([(0∧Q)≡R]'V[(0→R)∨(Q→R)]). Since ((0→R)≡1) [equivalently, (0'∨R)=1] we have that
([(0∧Q)≡R]'V[(0→R)∨(Q→R)])=([(0∧Q)≡R]'V[1∨(Q→R)])=([(0∧Q)≡R]'V1)=1
Case 2: Suppose P=1. Since (1∧Q)=Q, and (1→R)=R [equivalently, (1'∨R)=R] we have that
([(1∧Q)≡R]'V[(1→R)∨(Q→R)])=([Q≡R]'V[R∨(Q→R)])
Suppose Q=0 (case 2.1). We then have that
([0≡R]'V[R∨(0→R)])=([0≡R]'V[R∨1])=([0≡R]'V1)=1.
Suppose Q=1 (case 2.2). We then have that
([1≡R]'V[R∨(1→R)])=([1≡R]'V[R∨R])=([1≡R]'VR)
Since (1≡R)=R, we then obtain
([1≡R]'VR)=(R'VR)=1
Since P either equals 0, or 1 in the two-element Boolean Algebra {0, 1}, it follows that ([(P∧Q)≡R]'V[(P→R)∨(Q→R)]) equals 1 in the two-element Boolean Algebra. By the metatheorem mentioned above, it follows that ([(P∧Q)≡R]'V[(P→R)∨(Q→R)]) or equivalently ([(P∧Q)≡R]→[(P→R)∨(Q→R)]) holds for any Boolean Algebra.