Note that, when $D$ is diagonal:
$$(DA)_{ii} = D_{ii} A_{ii}$$
So $tr(DA) = \sum_{i=1}^n D_{ii} A_{ii}$. About the best bound you can do for this is the Cauchy-Schwarz inequality, i.e.
$$|tr(DA)| \leq \left ( \sum_{i=1}^n D_{ii}^2 \right )^{1/2} \left ( \sum_{i=1}^n A_{ii}^2 \right )^{1/2}$$
If you want a result in terms of traces, you can use the fact that $\| x \|_2 \leq \| x \|_1$ to get
$$|tr(DA)| \leq tr(|D|) tr(|A|)$$
where $(|D|)_{ij} = |D_{ij}|$ and similar for $|A|$. The first bound is attained when the diagonals of $D$ and $A$ are proportional to each other. The second is attained when these diagonals only have one nonzero entry. So for example you could have $D=A=\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}$, then $|tr(DA)|=1=1 \cdot 1 = tr(|D|) tr(|A|)$.
Note that when $D$ and $A$ both have nonnegative diagonal entries, we get the nice result
$$tr(DA) \leq tr(D) tr(A)$$
which is probably more like what you were looking for.
The answer to the body of your question is much quicker than the answer to the title.
Note that for any vector $x$, we have
$$
Ax = vv^Tx = v\langle x,v \rangle = \langle x,v \rangle v
$$
By definition, this is the projection of $x$ onto the vector $v$.
Yes, we could prove that in general, a matrix is an orthogonal projection if it is idempotent and symmetric. However, doing so is not necessary in answering this particular question.
Best Answer
If $p=0$ and $p=n$ this is trivial, since there is only one idempotent of rank $p$.
Claim: Let $1\le p< n$ be integers. The range of $M\longmapsto\mathrm{Tr}(M^TM)$ is equal to $[p,+\infty)$ when $M$ runs over the set of all $n\times n$ real idempotents of rank $p$.
Sketch of proof: