 Solution

The embedded $SU(3)$ acts on the first three coordinates and fixes the fourth. This immediately gives the branching rulesDistribute the tensor product of group representations over these direct sums and split $3\otimes \bar{3}=8\oplus 1$ by taking the trace:One can locate every summand using the block matrixThe traceless part of $A$ is $8$, $b$ is $3$, $c$ is $\bar{3}$, and $\mathrm{tr}A,d$ are independent trivial representations. Removing the overall trace instead gives $15\downarrow SU(3)=8\oplus 3\oplus \bar{3}\oplus 1$. The dimension check is $16=8+3+3+1+1$.