A function is self dual if it is equal to its dual (A dual function is obtained by interchanging $.$ and $+$).

For self-dual functions,

- Number of min terms equals number of max terms
- Function should not contain two complementary minterms - whose sum equals $2^{n}-1$, where $n$ is the number of variables.

$${\begin{array}{|c|c|c|c|}\hline

\textbf{}& \textbf{A}& \textbf{B}&\bf{C} \\\hline

0&0&0&0 \\1& 0&0&1 \\ 2& 0&1&0 \\ 3& 0&1&1 \\ 4& 1&0&0 \\ 5&1&0&1 \\ 6& 1&1&0 \\ 7&1&1&1\\ \hline

\end{array}}$$

So, here $(0,7) (1,6) (2,5) (3,4)$ are complementary terms so in self-dual we can select any one of them but not both.

Totally $2\times 2\times 2\times 2 =2^4$ possibility because say from $(0,7)$ we can pick anyone in minterm but not both.

For example, let $f = \sum (0,6,2,3)$

**NOTE: here I have taken only one of the complementary term for min term from the sets.**

So, remaining numbers will go to MAXTERMS

For above example, $2^4 =16$ self dual functions are possible

So, if we have $N$ variables, total Minterms possible is $2^n$

Then half of them we selected so $2^{n-1}$.

Now we have 2 choices for every pair for being selected.

So total such choices $=\underbrace{2\times 2\times 2\times 2\dots 2}_{2^{n-1}\text{ times} }$

$\therefore 2^{2^{n−1}}$ (option D)