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