1. Use determinant properties for adjugate matrices

For an $n \times n$ matrix $M$,

$$\det(\operatorname{adj}(M)) = (\det M)^{n-1}.$$

Here the matrix is $3 \times 3$, so

$$\det(\operatorname{adj}(M)) = (\det M)^2.$$

Apply this twice:

If $N = \operatorname{adj}(3A)$, then

$$\det(\operatorname{adj}(\operatorname{adj}(3A))) = (\det(\operatorname{adj}(3A)))^2.$$

Also,

$$\det(\operatorname{adj}(3A)) = (\det(3A))^2.$$

Hence,

$$\det(\operatorname{adj}(\operatorname{adj}(3A))) = ((\det(3A))^2)^2 = (\det(3A))^4.$$

So we only need $\det(3A)$.

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2. First compute $\det(A)$

Given

$$A = \begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix}.$$

Let the columns be $C_1, C_2, C_3$.

Then

$$C_1 = \begin{bmatrix}2\\4\\6\end{bmatrix}, \quad C_2 = \begin{bmatrix}2+p\\6+2p\\12+3p\end{bmatrix}, \quad C_3 = \begin{bmatrix}2+p+q\\8+3p+2q\\20+6p+3q\end{bmatrix}.$$

Now perform column operations that do not change determinant:

$$C_2 \to C_2 - C_1, \qquad C_3 \to C_3 - C_2.$$

Compute them:

$$C_2 - C_1 = \begin{bmatrix}p\\2+2p\\6+3p\end{bmatrix},$$ $$C_3 - C_2 = \begin{bmatrix}q\\2+p+2q\\8+3p+3q\end{bmatrix}.$$

So

$$\det(A)= \begin{vmatrix} 2 & p & q \\ 4 & 2+2p & 2+p+2q \\ 6 & 6+3p & 8+3p+3q \end{vmatrix}.$$

Now do row operations:

$$R_2 \to R_2 - 2R_1, \qquad R_3 \to R_3 - 3R_1.$$

Then

$$R_2 = [0,2,2+p], \qquad R_3 = [0,6,8+3q].$$

Thus

$$\det(A)= \begin{vmatrix} 2 & p & q \\ 0 & 2 & 2+p \\ 0 & 6 & 8+3q \end{vmatrix}.$$

Expand along the first column:

$$\det(A)=2\begin{vmatrix}2 & 2+p\\6 & 8+3q\end{vmatrix}.$$

Now,

$$\begin{vmatrix}2 & 2+p\\6 & 8+3q\end{vmatrix} =2(8+3q)-6(2+p) =16+6q-12-6p =4+6q-6p.$$

So,

$$\det(A)=2(4+6q-6p)=8+12q-12p.$$

But this still depends on $p,q$, which suggests we should re-check the simplification carefully.

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3. Better simplification using row structure

Let us instead use row operations directly on the original matrix:

$$R_2 \to R_2 - 2R_1, \qquad R_3 \to R_3 - 3R_1.$$

From the original matrix,

$$R_2 - 2R_1 = [4,6+2p,8+3p+2q] - 2[2,2+p,2+p+q]$$ $$= [0,2,p+4],$$

and

$$R_3 - 3R_1 = [6,12+3p,20+6p+3q] - 3[2,2+p,2+p+q]$$ $$= [0,6,14+3p].$$

Hence,

$$\det(A)= \begin{vmatrix} 2 & 2+p & 2+p+q \\ 0 & 2 & p+4 \\ 0 & 6 & 14+3p \end{vmatrix}.$$

Expand along first column:

$$\det(A)=2\begin{vmatrix}2 & p+4\\6 & 14+3p\end{vmatrix}.$$

Now,

$$\begin{vmatrix}2 & p+4\\6 & 14+3p\end{vmatrix} =2(14+3p)-6(p+4) =28+6p-6p-24=4.$$

Therefore,

$$\det(A)=2\cdot 4=8.$$

So,

$$\det(3A)=3^3\det(A)=27\cdot 8=216=2^3\cdot 3^3.$$

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4. Now compute the required determinant

As found earlier,

$$\det(\operatorname{adj}(\operatorname{adj}(3A)))=(\det(3A))^4.$$

Thus,

$$(216)^4=(2^3\cdot 3^3)^4=2^{12}\cdot 3^{12}.$$

So,

$$m=12,\qquad n=12.$$

Hence,

$$m+n=24.$$

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5. Check with options

The correct option is:

$$\boxed{24}$$

which is Option C.