Time complexity for $B(n), T_B(n) = T_B(n-1) + 1 \implies T_B(n) = \Theta(n)$
Time complexity for $D(n) = \Theta(1)$
So, time complexity for $A(n), T_A(n) = 2T_A\left(\sqrt n\right) + \Theta(n)$
For sake of simplicity rewriting $T_A$ as $S,$ we get
$S(n) = 2S(\sqrt n)+ \Theta(n)$
Substituting $n = 2^m$
$S(2^m) = 2S(2^{m/2}) + \Theta (2^m)$
Now taking $S(2^m) = R(m)$
$R(m) = 2R(m/2)+ \Theta(2^m)$
On this we can apply Master's theorem case 3, and we get $R(m) = \Theta (2^m)$
$\implies S(2^m) = \Theta(2^m)$
$\implies S(n) = \Theta(n)$