The language $L$ has three distinct language quotients, corresponding to languages accepted from the three states of the minimal DFA:
$$\begin{align}
(\Sigma^*bb)^{-1} L &= \{\, w \mid bbw \in L\,\} = L \setminus \{\, \epsilon, b \,\} \\
(\epsilon \cup \Sigma^*a)^{-1} L &= \{\, w \mid aw \in L\,\} = L \\
(b \cup \Sigma^*ab)^{-1}L &= \{\, w \mid abw \in L\,\} = L \setminus \{\, b\,\} \enspace,
\end{align}$$
where $\Sigma = \{\,a,b\,\}$. A DFA must have at least one state per distinct quotient (exactly one if the DFA is minimal). In general, an NFA may have fewer states than quotients, because a quotient may equal the union of languages accepted from multiple states.
However, in this case it's not possible to go below three states, because there must be two distinct accepting states and one non-accepting state. In fact, both $(\epsilon \cup \Sigma^*a)$ and $(b \cup \Sigma^*ab)$ are subsets of $L$, but have distinct quotients and cannot be merged.
The non-accepting state is necessary because the quotient of $(\Sigma^*bb)$ is nonempty. Hence there must be a non-accepting state with a path to accepting states.
Once you fix the three states, you also fix the languages accepted from them. This means that in this case all transitions in the DFA are essential. We can add some redundancy, though. Observe that
$$ L \setminus \{\,\epsilon,b\,\} \subseteq L \setminus \{\,b\,\} \subseteq L \enspace. $$
If there is a transition on input $\sigma$ from $q$ to $q'$, we can freely add a transition from $q$ to $q''$ for the same input as long as the language accepted from $q''$ is a subset of the language accepted from $q'$.
For example, we can add a transition labeled $b$ from $q_0$ to $q_2$. Of course, if we count states and transitions, our automaton hasn't gotten any simpler.
