What would NP-complete solution in O(2^N/B) mean?
It says nothing about the P vs NP question because no such algorithm can exist.  If there is an algorithm that takes as its input a formula $\phi$ and an integer&nbsp;$B$ and determines whether $\phi$&nbsp;is satisfiable in time $O(2^{N/B})$, where $N$&nbsp;is the length of&nbsp;$\phi$, then we can decide SAT in constant time by calling the algorithm with $B=N$.

But we know that SAT cannot be solved in constant time because, in $k$&nbsp;steps of the computation, you can only read the first $k$&nbsp;characters of the input. That means that you have no way of distinguishing between the formulas $X\land X\land\dots \land X$ and $X\land X\land\dots \land X\land\neg X$, where $X$&nbsp;is repeated, say, $k$&nbsp;times. One of these formulas is satisfiable and the other isn't, but an algorithm running in $k$&nbsp;steps would have to return "satisfiable" or "unsatisfiable" before seeing the $\neg X$ (or lack of it) at the end.