When source move towards observer w.r.t. medium, distance between two consecutive wave - front decreases and there by decreasing time taken in travelling from one wave front to other and hence increasing frequency. In case of 'A' source moves towards observer w.r.t. medium white in B and C source moves away from observer . Where as in D source doesn't move w.r.t. no medium.
Alternative :

(A)    (Q,T)

If AA' and BB' are considered to be fixed planes then            
Source moves towards right
v - speed of sound
d - remains same
v - remains same

$\mathrm{t}=\frac{\mathrm{d}}{\mathrm{v}}=\text{ remains the same }$

(A)    (Q,S,T)

If AA' and BB' are considered as wave fronts if sources moves toward right

$\mathrm{\lambda }=\frac{\mathrm{v}}{\mathrm{f}}$

${\mathrm{f}}^{\text{'}}=\frac{\mathrm{v}}{{\mathrm{\lambda }}^{\text{'}}} {\mathrm{\lambda }}^{\text{'}}=\frac{\mathrm{vt}-{\mathrm{v}}_{\mathrm{s}}\mathrm{t}}{\mathrm{ft}}=\left(\frac{\mathrm{v}-{\mathrm{v}}_{\mathrm{s}}}{\mathrm{f}}\right)$

${\mathrm{f}}^{\text{'}}=\mathrm{f}\left(\frac{\mathrm{V}}{\mathrm{V}-{\mathrm{v}}_{\mathrm{S}}}\right)$

f ' > f
Since distance between AA' and BB' wavefront decreases hence time taken by soun wave decreases.
(B)    (P,S)
If AA' and BB' are considered as fixed planes. Air moves toward right

${\mathrm{t}}^{\text{'}}=\frac{\mathrm{d}}{\mathrm{v}+{\mathrm{v}}_{\text{air }}}\equiv \text{ decreases }$

(B)    (P)
If AA' and BB' are considered as wave fronts distance between then increases.

$\text{ Initially }\mathrm{t}=\frac{\mathrm{d}}{\mathrm{v}}=\frac{\mathrm{\lambda }}{\mathrm{v}}=\frac{1}{\mathrm{f}}$

${\mathrm{x}}^{\text{'}}=\frac{\mathrm{v}+{\mathrm{v}}_{\text{air }}}{\mathrm{f}}\left(\mathrm{v}+{\mathrm{v}}_{\text{air }}\right)={\mathrm{v}}_{\text{net }}\left(\text{ net velocity of sound }\right)$

$\frac{\left(\mathrm{v}+{\mathrm{v}}_{\text{air }}\right)}{\mathrm{f}(\mathrm{v}+\mathrm{air})}={\mathrm{t}}^{\text{'}}=\frac{{\mathrm{d}}^{\text{'}}}{\mathrm{v}+{\mathrm{v}}_{\text{air }}}\equiv \text{ remains same }$

${\mathrm{t}}^{\text{'}}=\frac{1}{\mathrm{f}}$

(C)    (P)
If AA' and BB' are considered to be fixed planes and source and observer 
move toward left.                                    
d - remains same

$\mathrm{t}=\frac{\mathrm{d}}{\mathrm{v}}\equiv \text{ remain same }$

(C)    (P,R)

If AA' and BB' are considered wavefronts, distance between them increase and hence time taken by sound increases.

(D)    (S,T)

If AA' and BB' are considered fixed planes source and medium both move toward right dut to motion of medium 

t = d/v
d → remains same v_net = v + v_air (net velocity of sound increases)

${\mathrm{t}}^{\text{'}}=\frac{\mathrm{d}}{\left(\mathrm{v}+{\mathrm{v}}_{\text{air }}\right)}=\text{ decreases }$

(D)    (S,T)

If AA' and BB' are considered wavefronts.

Due to motion of source toward right distance between wavefront decreases and hence time taken by sound decreases.