Since intuitively you can't make player 1 (rows player) indifferent between choosing $C$ or $D$ (he is always better-off choosing $D$), we should expect p=1
Actually, since player I can't be made indifferent between $C$ and $D$ (he always prefers $D$), there is no solution to the equation $u_1(C)=u_1(D)$. This is exactly what you have discovered.
In general, if player I's payoff matrix is $A$, then against a mixed strategy $y$ of player II, represented as a column vector, player I expects payoff $(Ay)_i$ for row $i$. For a given subset of rows, $R$, we may or may not be able to find $y$ such that $(Ay)_i = u$ for all $i \in R$.
In fact, even if we can make player I indifferent between the rows in $R$ such that he expects payoff $u$, it could still be the case that there is a row $j \notin R$ with $(Ay)_j > u$, i.e., against $y$ the rows in $R$ are not best responses.
Here's a $3 \times 2$ example:
$A = \left ( \begin{array}{cc} 0 & 2\\ 2 & 1\\ 3 & 0 \end{array} \right )$
For $y = (1/3, 2/3)^\top$ we have $Ay = \left ( \begin{array}{c}{4/3\\4/3\\1}\end{array} \right )$, so rows 1 and 2 are best responses against $y$. Likewise, for $y = (1/2, 1/2)^\top$, rows 2 and 3 are best responses with expected payoff 1.5 versus payoff 1 for row 1. However, for $y = (2/5, 3/5)^\top$, player I is indifferent between rows 1 and 3, which have payoff $6/5$, but the unique best response is row 2 with payoff $7/5$.
In general, potential Nash equilibria correspond to vertices of polyhedra defined by an upper envelope of payoff hyperplanes for pure strategies against the mixed strategy of the opponent. Some sets of hyperplanes do not meet in a vertex, and some that do meet in a vertex are below the upper envelope, i.e., they are not best responses against the mixed strategy that equalizes the payoffs of the corresponding pure strategies.
For more details on finding equilibria using upper envelopes, see the answer to this question:
Mixed strategy nash equilibria in from $2\times N$ bimatrix form
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
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."