Circuit Using only NAND and NOT Gates
You really cannot simplify \$A + \bar B + D\$.  

But \$ \overline {A C} + \bar C D\$ can be simplified.  Take deMorgan's on first term.

\$ \bar A + \bar C + \bar C D = \bar A + \bar C ( 1 + D ) = \bar A + \bar C = \overline {A C}\$

No possibility of combining the two circuits.  

So using NANDs as NOTs, you have 6 NANDs total.