Let's call player's $1$ strategies A and B, and player's $2$ strategies C and D. Let C be the dominant one. So, the payoff matrix is:
$$\small \begin{matrix}
& & \textrm{player } 1&\\
& & \textrm{A} & \textrm{B} \\
\textrm{player } 2 &\textrm{C} & x_1, x_2 & y_1, y_2\\
& \textrm{D} & u_1, u_2 & z_1, z_2 \\
\end{matrix}$$
Since A is dominant, $x_2>y_2$ and $u_2 > z_2$.
Since player $1$ always plays A we can ommit strategy B and the game's matrix actually looks like this:
$$\small \begin{matrix}
& & \textrm{player } 1&\\
& & \textrm{A} \\
\textrm{player } 2 &\textrm{C} & x_1, x_2\\
& \textrm{D} & u_1, u_2 \\
\end{matrix}$$
Assuming that player $2$ goes only for maximising her own profit (i.e. she doesn't get any happier when player $1$ gets less), player $2$ would play the strategy that gives her most profit, that is:
if $x_1>u_1$ player $2$ plays A, if $x_1<u_1$ player $2$ plays B.
If $x_1 = u_1$ then there are continuum different mixed stategy NEs since any pair $\{(1,0), (t, (1-t))\}$for $t\in[0,1]$ gives a NE.
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."
Best Answer
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.