You got most of the way. Follow what's below:
$$\begin{align*}
F &= \overline{(C + \overline{D})}\: \overline{(B + D)} + \overline{\overline{(C + \overline{D})}}\: \overline{\overline{(B + D)}}\tag{0}\\\\
&=\overline{C} \:D\: \overline{B}\: \overline{D} + (C + \overline{D})\: (B + D)\tag{1}\\\\
&=(C + \overline{D})\: (B + D)\tag{2}\\\\
&=B\:C + C\: D + B\:\overline{D} + D\:\overline{D}\tag{3}\\\\
&=B\:C + C\: D + B\:\overline{D}\tag{4}\\\\
&=B\:C\:D + B\:C\:\overline{D} + C\: D + B\:\overline{D}\tag{5}\\\\
&=(B\:C + C)\: D+(B\:C + B)\:\overline{D}\tag{6}\\\\
&=(C\:[B + 1])\: D+(B\:[C + 1])\:\overline{D}\tag{7}\\\\
&=(C\cdot1)\: D+(B\cdot1)\:\overline{D}\tag{8}\\\\
&=C\: D+B\:\overline{D}\tag{9}
\end{align*}$$
Here, you can see that I've expanded the BC term in step 5. This is just turning a single case into two cases, which does NOT change the result. I think you can see that it doesn't, by inspection.
Then, in step 6 I organize the summed terms so that I can factor out D and Not-D, to create two somewhat more complex terms. But now, simple inspection tells you that in the first term (based on D) that if C is true that it doesn't matter whether or not B C is true and that if C is false then so is B C. So that can be reduced down to just C. The same idea also applies to the second term (based on Not-D.) I've added steps 7 and 8 to show this transition.
The final result is in 9, now.
