The correct option is A only II and IV
Conditions for geometrical isomers:
1) Free rotation must be restricted.
2) Both the terminals carbon must be attached with different groups.
3) Terminals must not be in perpendicular planes.
II and IV follow all the above conditions. So these two are pair of geometrical isomers.
Structure III is a structural isomer of II and IV.
Structure I cannot show geometrical isomer because two same group (Br) is attached with one of the terminal carbon of double bond. Hence the correct option is a.