What is the worst-case runtime complexity to transform a NFA to DFA via Rabin-Scott's power set construction?
At first glance this looks like a mistake in Wikipedia (it wouldn't be the first one), but I think the time bound there is actually correct, with two assumptions and one small idea.

The assumption is: the alphabet size is constant, so you can ignore the $b$ factor in the runtime and output size. The second assumption is: you can represent the output DFA as a graph in memory, in a conventional adjacency list data structure (or whatever). So in particular to be able to store an edge from one state to another, you need memory cells that can store up to $n$ bits of information per cell, and (in the standard RAM model) this also means constant-time bitwise operations on these cells.

The idea is: each nonempty set of states of the NFA (corresponding to a single state of the DFA) can be formed by adding one element to a smaller set. So, to compute the transitions for a state $S$: find an element $s$ and a state $R$ such that $S=R\cup\{s\}$. Then, for each input alphabet symbol, compute the transition from $S$ as the bitwise Boolean or of the transition from $R$ (previously computed in the same construction) and the transition from $s$ (in the input NFA). Doing this takes $O(b)$ bitwise Boolean operations per state for a total time of $O(1)$ per state.

In practice, you'd probably rather use the version of the construction that only constructs reachable states, rather than all sets. But when you do it that way you can't be guaranteed to find an element whose removal leaves another reachable state, so your $n$ extra factor in the runtime presumably still applies to this version.