Equivalence classes of A form a partition of A. Option C
Let's get into details:-
For the concept of equivalence classes to even be mentioned, the relation must be equivalence first. An equivalence relation is a relation that is reflexive, symmetric and transitive — all the three.
In contrast, a relation that is reflexive, anti-symmetric and transitive is a partial order.
Let's take a set $A=\left \{ 1,2,3,4 \right \}$
Let's define a relation R on A. $R=\left \{ (1,1),(2,2),(3,3),(4,4),(1,2),(2,1) \right \}$
Clearly, R is equivalence.
To find equivalence classes of each element, we find what element is our element personally related to.
(x,y) means x is related to y NOT y is related to x.
1 is related to 1 and 2. So, $\left \{ 1,2 \right \}$
2 is related to 2 and 1. So, $\left \{ 1,2 \right \}$
3 is related to 3. So, $\left \{ 3 \right \}$
4 is related to 4. $\left \{ 4 \right \}$
5 is related to 5. $\left \{ 5 \right \}$
These are all the equivalence classes, and hence the partitions of A.
PS: Remember that union of all equivalence classes is the original set, and intersection of all is $\phi$