Let $$P[P_{ij}]$$ and $$Q=[q_{ij}]$$ be two square matrices of order 3 such that $$q_{ij}= 2^{(i+j-1)}p_{ij}$$ and $$\det (Q)=2^{10}.$$ Then the value of det(adj(adj P)) is:

Given $$q_{ij} = 2^{i+j-1} p_{ij}$$. In a $$3 \times 3$$ matrix, this means:

$$Q = \begin{bmatrix} 2^1 p_{11} & 2^2 p_{12} & 2^3 p_{13} \\ 2^2 p_{21} & 2^3 p_{22} & 2^4 p_{23} \\ 2^3 p_{31} & 2^4 p_{32} & 2^5 p_{33} \end{bmatrix}$$

Factor $$2^1, 2^2, 2^3$$ from rows and $$2^0, 2^1, 2^2$$ from columns:

$$\det(Q) = (2^1 \cdot 2^2 \cdot 2^3) \cdot (2^0 \cdot 2^1 \cdot 2^2) \cdot \det(P) = 2^6 \cdot 2^3 \cdot \det(P) = 2^9 \det(P)$$.

$$2^{10} = 2^9 \det(P) \implies \det(P) = 2$$.

Final Calculation: $$\det(adj(adj P)) = |P|^{(3-1)^2} = |P|^4 = 2^4 = 16$$.

Correct Option: D