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Let L be a lattice and A be any subset of L with two minimal proper supersets B and C in L. My doubt is that whether B and C would be isomorphic .I got some special cases in which they are isomorphic .Is it true always.

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    Maybe you should elaborate a little more in what you mean. A minimal superset of $A$ is $A$ itself, and the only one. Maybe you mean $B$ and $C$ are minimal proper supersets of $A$ in $L$? Or would they minimal supersets of $A$ among the sub-lattices of $L$? – 2017-02-24

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The answer is no.

Here's a counterexample. (If you don't understand the notation, then I'll draw the lattice.)

Let $\mathbf{L}$ be the lattice $\mathbf{1} \oplus \mathbf{2}^2 \oplus \mathbf{1}$, where $\mathbf{1}$ is the (unique, up to isomorphism) one-element lattice, $\mathbf{2}$ is the (unique, up to isomorphism) two-element lattice, and $\oplus$ is ordinal sum of posets, consisting in putting the poset on the right (second operand) on top of the left one, and the exponentiation is the direct power of posets.
So $\mathbf{2}^2$ is the (unique, up to isomorphism) four-element lattice which is not a chain, and $\mathbf{L}$ is that lattice with one bottom and one top element added.
Let us say the elements of $\mathbf{L}$ are $\{ \bot, 0, a, b, 1, \top \}$, with the covering relations $$\bot \prec 0, \, 0 \prec a, \, 0 \prec b,\, a \prec 1,\, b \prec 1, 1 \prec \top. $$

Now let $A$ consist of the four "middle" elements: $A = \{ 0, a, b, 1\}$, $B = \{\bot\} \cup A$, and $C = A \cup \{\top\}$.
Then $B$ and $C$ are minimal proper supersets of $A$ within $L$, and although they're both lattices, they're not isomorphic (they're dually isomorphic).
Here I am assuming that by isomorphic you don't just mean that they have the same cardinal.