It's like tonic, which isn't considered a mixed drink ($0$ parts gin and $1$ part tonic). That's degenerate for you: "In mathematics [as in mixology], a degenerate case is a limiting case in which a class of object changes its nature so as to belong to another, usually simpler, class."
I think the point of this problem is to illustrate the difference between a Nash equilibrium and a subgame-perfect Nash equilibrium.
For subgame-perfect equilibria, the situation is clear. Player $2$ cannot make an empty threat to play $(0,0)$ in the left-hand branch; she plays $(5,5)$, and thus Player $1$ plays $A$; there is no other equilibrium.
If the Nash equilibrium doesn't have to be subgame-perfect, the situation is quite a bit more complicated. Player $2$ can threaten to play $(0,0)$ with non-zero probability $1-p$ in the left branch, thus making this branch worth an expected payoff of $5p$ for Player $1$. In the right-hand branch, as you wrote, Player $1$ will always play "No", and Player $2$ can play L with probability $q$ and NL with probability $1-q$, giving an expected payoff of $3q+4(1-q)=4-q$ to Player $1$.
There are two types of equilibria, depending on whether Player $1$ prefers A or not.
Player $1$ prefers A: Here Player $2$ must choose $p=1$, since the left-hand branch actually gets played. $q$ can be anything, since the right-hand branch doesn't get played and any value of $q$ lets Player $1$ prefer $A$ if $p=1$.
Player $1$ doesn't prefer A: Here Player $2$ can have mixed strategies in both branches, subject to the condition $4-q\ge5p$ so that Player $1$ doesn't prefer A. In these equilibria, Player $1$ always plays B. Note that this includes (weak) equilibria with $4-q=5p$ in which Player $1$ is indifferent between A and B. Nevertheless, for an equilibrium Player $1$ must play B and cannot mix strategies, since Player $2$'s empty threat in the left branch wouldn't be in equilibrium otherwise.
Best Answer
The idea is that if all the other players strategies are unchanged, and I have the strategy X (different from chosing 0 everytime), the strategy "chosing the number given by the strategy X minus one (or 0 if the strategy X gives me 0)" is always better.
Hence, the strategy X can be improved, and we're not in a Nash equilibrium