Above is a problem from my end sem mathematics question paper.
In this question I have two basic problems or else I have not understood the question properly
Notation A/R is used for quotient set(till now I only know this), each element of a quotient set is itself a set. Now it is given that A/R = P, if A is a set whose elements are not sets then P being a subset of A will also have elements which are not set, which contradicts the property of a quotient set. Basically I have problem in understanding what is the meaning of A/R=P?
Let us ignore the notion as asked above. I assume that A/R=P means that the quotient set contains the equivalence class of elements belonging to set P.
Let A={1,2,3} and let P={2,3}
There is a theorem which states that if we partition any non-empty A, then we can define an equivalence relation on A such that each partition will be an equivalence class. So now clearly the A/R will contain the equivalence class of element '1' then A/R $\ne$ P. This means that there exists no such R. This again shows that the question is wrong.
Above are the two confusions, please check it and clear my doubts. If I'm understanding the question wrong then please make me understand the question clearly.
Thank you!

For every equivalence relation $R$ on $A$, the quotient $A/R$ is the set of $R$-equivalance classes $[x]$ of elements $x \in A$. Here, $[x]=\{y\in A\,|\,yRx\}$ is the set of elements of $A$ that are equivalent to $x$ (under $R$).
Now, a partition $P$ of a set $A$ is a collection of disjoint subsets $\mathcal{A}_k$ of $A$ whose union is $A$. In symbols, $P=\{\mathcal{A}_k\,|\,k\in I\}$, $\mathcal{A}_j\cap\mathcal{A}_k=\emptyset$ whenever $j\neq k$ and $\cup_{k\in I}\mathcal{A}_k=A$. Here, $I$ is some arbitrary index set.
When one writes $A/R=P$ it means that each equivalence class $[x] \in A/R$ corresponds to one of the $\mathcal{A}_k$'s in $P$.
Can you think of something that relates equivalence relations and partitions?