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."
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
Best Answer
Suppose Player 1 chooses uniformly a bid in the set $\{0,\ldots,99\}$ and denote it by $X$. When Player 2 bids $i$, his payoff is $100\Pr(X< i) -i +50\Pr(X=i)$, where I assumed that when both bid the same, they split the $100$. Since $\Pr(X<i)=\tfrac{i}{100}$, the payoff is $50\Pr(X=i)=0.5$ it is independent of $i$. Thus, Player 2 is indifferent among all the bids. This is true when player 2 uses this strategy too, so when both chose a number uniformly, neither has a profitable deviation and it's an equilibrium.
Note that the $50$ doesn't really matter. Even if they don't split the $100$ when bid the same but each gets $100$, the indifference holds. However, there might be other pure equilibria in this case (each strategy profile where they bid the same is a pure equilibrium in this case).