Game with an equilibrium in pure, but none in mixed strategies?
As I said in the comment, Nash's theorem shows the existence of a Nash equilibrium (possibly but not necessarily in mixed strategies). If you are interested in Nash equilibria is proper mixed strategies, i.e., NE in which all players play at least two actions with positive probability, you can easily show the following impossibility result: No such NE can exist if one player $i$ has a stricly dominant strategy, i.e., a strategy that gives a higher payoff regardless of the other players' strategy. Since the strictly dominant strategy is $i$'s unique best response against all strategies of the others, this pure strategy must be a part of each equilibrium. Each other strategy by $i$ can be improved upon by deviating to the dominant strategy.
In your case 1 has a strictly dominant strategy, $B$ - so even if 2 always plays $L$, 1 would not be indifferent between $A$ and $B$, but always choose $B, 9&gt;8$. Without indifference, mixing is never optimal and therefore never part of a NE.
Even if $B$ was only weakly dominant, it would always be the unique best response to every strategy in which player 2 plays $L$ and $N$ with positive probability.