Given an non-deterministic finite automaton, will its determinization always have unreachable states?
No, there aren't always unreachable states.  Consider the NFA with one state, $q$, and no transitions.  (It accepts the language $\{\epsilon\}$ if $q$&nbsp;is accepting, and accepts&nbsp;$\emptyset$, otherwise.)

If you determinize this automaton, you end up with a two-state DFA with a transition from the start state $\{q\}$ to the other state, $\emptyset$, so both states are reachable.