Whenever you see a matrix trace, you should think inner product, because
$$ \operatorname{Tr}(A^T B) = \langle A, B\rangle_F = \langle A,B\rangle_{\mathbb R^m \otimes \mathbb R^n}$$
that is, the trace of the product of two matrices is equal to their frobenius inner product, which in turn is the induced inner product on the tensor product of Hilbert spaces.
Since it is an inner product, the Cauchy-Schwartz inequality applies:
$$ |\langle A, B \rangle_F |^2 \le \|A\|_F^2\|B\|_F^2$$
with equality if and only if $A$ and $B$ are linearly dependent matrices, i.e. scaler multiples of each other. In your case, we have
$$ |\operatorname{Tr}(ABA^T)| = |\operatorname{Tr}(A^TA B)| = |\langle A^TA , B\rangle_F| \le \|A^TA\|_F\|B\|_F$$
The last term can be further bounded by
$$\begin{aligned}
\|A^TA\|_F\|B\|_F &\le \|A\|_F^2\|B\|_F = \Big(\sum\nolimits_i \sigma_i^2(A)\Big)\cdot\sqrt{\sum\nolimits_j \sigma_j^2(B)}
\\ &\le rank(A)\cdot \sigma^2_{\max}(A)\cdot rank(B)\cdot\sigma_{\max}(B)\\
&\le m\cdot \min(m,n)\cdot \sigma^2_{\max}(A)\cdot \sigma_{\max}(B)
\end{aligned}$$
Best Answer
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.