For a dibasic acid, $H_{2}AH{A}^{-}+H^{+}\left(K_1\right)$
$H{A}^{-}{A}^{2-}+H^{+}\left(K_2\right)$;
$H_{2}A2H^{ }+A^{2-}\left(K\right)$ then,

$K_1=\frac{\left[HA\right]\left[H^{+}\right]}{\left[H_{2}A\right]}$
For reaction 2;
$K_2=\frac{\left[A^{2-}\right]\left[H^{+}\right]}{\left[HA\right]}$
For reaction 3;
$K=\frac{\left[A^{2-}\right]{\left[H^{+}\right]}^2}{\left[H_{2}A\right]}$
On solving this further for $K=K_1.K_2$
$K=K_1.K_2$
   =  $\frac{\left[HA\right]\left[H^{+}\right]}{\left[H_{2}A\right]}$ X$\frac{\left[A^{2-}\right]\left[H^{+}\right]}{\left[HA\right]}$
   =   $\frac{\left[A^{2-}\right]{\left[H^{+}\right]}^2}{\left[H_{2}A\right]}$
Hence, the correct answer is option (D)$K=K_1.K_2$.