I have the following boolean logic: $$ \overline {\overline {\overline {B+C+D} + \overline {DA}} + \overline {\overline {\overline {A+E} + \overline { B}} + \overline {E}}} $$
I am trying to simplify the logic, what confuses me is why I cannot apply De Morgan's Law $\overline {A + B} = \overline {A} \cdot \overline {B}$.
In this case: $$ \overline {\overline {\overline {B+C+D} + \overline {DA}}} \cdot \overline {\overline {\overline {\overline {A+E} + \overline { B}} + \overline {E}}} $$
You are correct.
In general, De Morgan's Law can be applied as needed to simplify the expression.
In this case, it is best to start with the full expression as you have done.
$$\begin{align} \overline {\overline {\overline {B+C+D} + \overline {DA}} + \overline {\overline {\overline {A+E} + \overline { B}} + \overline {E}}} \\[2ex] \left(\overline {\overline {\overline {B+C+D} + \overline {DA}}}\right) \cdot \left(\overline {\overline {\overline {\overline {A+E} + \overline { B}} + \overline {E}}}\right) \\[2ex] \left(\overline {B+C+D} + \overline {DA}\right) \cdot \left(\overline {\overline {A+E} + \overline { B}} + \overline {E}\right) \end{align}$$
Then apply De Morgan's Law to terms to get the sum of products and simplify.