Let DP = PQ = QB = \(x\)
Also, let the length of the rectangle be \(L\) and width be \(B\)

We know, diagonal of a rectangle = \(\sqrt{L^2 + B^2}\)
i.e. \(3x = \sqrt{L^2 + B^2}\)
\(L^2 + B^2 = 9x^2\)..........(1)

Now, In triangle APD;
\(B^2 = AP^2 + x^2\)
\(AP^2 = B^2 - x^2\)

And, In triangle APB;
\(L^2 = AP^2 + (2x)^2\)
\(AP^2 = L^2 - 4x^2\)

Equating \(AP^2\);
\(B^2 - x^2 = L^2 - 4x^2\)
\(L^2 - B^2 = 3x^2\).........(2)

Adding (1) and (2);
\(2L^2 = 12x^2\)
\(L^2 = 6x^2\)

So, \(B^2 = 3x^2\)

\(\frac{AB}{AD} = \frac{L}{B} = \sqrt{\frac{6x^2}{3x^2}} = \sqrt{2}\)

Hence, option A