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Set Theory

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In a group 
1] there exist a unique identity element of Group i.e G. lets call it ' e '
2] the inverse of any element in G is unique. We generally represent inverse of a element 'a' as $a^{-1}$

Now the equation is x*a = b

post multiply by $a^{-1}$ both side,

x*a*$a^{-1}$ = b * $a^{-1}$

x* e  = b * $a^{-1}$

x =  b * $a^{-1}$

$\because $ there always exit only unique inverse of any element in Group so solution to this equation is also unique.

Ref : https://en.wikipedia.org/wiki/Group_(mathematics)#Uniqueness_of_identity_element_and_inverses

This way answer is a) Unique Solution b * $a^{-1}$

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