= Solution
Choose <measurable sets> $S_i$ carrying $\mu_i^s$ with $\kappa(S_i)=0$, and put $S=\bigcup_iS_i$. Every absolutely continuous component vanishes on $S$, and every singular component vanishes on its complement. The given restriction identity therefore gives
$$
\mu|_S=\bigvee_i\mu_i^s,\qquad
\mu|_{S^c}=\bigvee_i\mu_i^a.
$$
The first is singular with respect to $\kappa$. The second is absolutely continuous because it is bounded above by $\sum_i\mu_i^a$. They sum to $\mu$, so uniqueness identifies them as its Lebesgue components.
Let $h_i=d\mu_i^a/d\kappa$ and $h=\max_i h_i$, using measurable representatives. Each partition sum is at most $\int_Ah\,d\kappa$. For the reverse inequality, partition $A$ measurably by assigning a point to the smallest index attaining this finite maximum. On its assigned cell $h_i=h$, and the partition sum is exactly $\int_Ah\,d\kappa$. Thus
$$
\boxed{\frac{d\mu^a}{d\kappa}=\max_{1\leq i\leq N}\frac{d\mu_i^a}{d\kappa}\quad\kappa\text{-almost everywhere}.}
$$
The maximum of finitely many integrable nonnegative densities is integrable, since it is at most their sum.
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