We are given $P=[a_{ij}], \qquad Q=[b_{ij}], \qquad b_{ij}=2^{i+j}a_{ij}, \quad 1\le i,j\le 3.$ Also, $\det(P)=2.$

Write the entries of $Q$ in a factorized form: $b_{ij}=2^{i+j}a_{ij}=2^i\,a_{ij}\,2^j.$

This means $Q$ can be written as $Q=D_1\,P\,D_2,$ where

Now use the determinant property: $\det(Q)=\det(D_1)\det(P)\det(D_2).$

Compute the determinants of the diagonal matrices: $\det(D_1)=2^1\cdot 2^2\cdot 2^3=2^{1+2+3}=2^6,$ and similarly, $\det(D_2)=2^6.$

Therefore, $\det(Q)=2^6\cdot 2 \cdot 2^6=2^{12}\cdot 2=2^{13}.$

So the determinant of $Q$ is $\boxed{2^{13}}.$

Comparing with the options:

• A: $2^{10}$
• B: $2^{11}$
• C: $2^{12}$
• D: $2^{13}$

Hence the correct option is D.