 Solution

Let $T$ commute with the given representation. Since $A$ and $C$ are diagonal and their four joint pairs of eigenvalues are distinct, commuting with both forces $T$ to be diagonal. Commuting with the permutation matrix $B$, which exchanges coordinates $1\leftrightarrow 2$ and $3\leftrightarrow 4$, equates the first two and last two diagonal entries. Commuting with $D$, which exchanges $1\leftrightarrow 3$ and $2\leftrightarrow 4$, makes all four entries equal. Thus the commutant consists only of scalar matrices. Since the representation is unitary, a proper invariant subspace would have an orthogonal projection in the commutant. No such nonscalar projection exists, so the representation is irreducible.