This is the best approach written by @NeelParekh (by converting into decimal).
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Another short & Intuitive approach:
A) We know, 2's complement addition overflow condition: $a_{n-1}b_{n-1}c_{n-1}^{'} + a_{n-1}^{'}b_{n-1}^{'}c_{n-1}$ i.e MSB of two no must be SAME. But here given two no's msb are diff, so no chance of Overflow.
Ref: GATE CSE 2006 | Question: 39
D) $2*B ⇒ \underbrace{\boxed{11010} \ 0}_\text{arithmatic left shift}⇒ \underbrace{\cancel{1} \boxed{10100}}_\text{considering 5 bits} ⇒ \underbrace{01100}_\text{2's complement} ⇒ -12 \ (\text{within range}) $
So, no chance of Over or Underflow.
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Now, as $A$ is a $+ve$ num & $B$ is $-ve$ num
so, $B-A = -ve - (+ve) = -ve $
and $A - B = +ve - (-ve) = +ve $
So there is a chance of these 2 cases we may face $\text{arithmatic Overflow or Underflow }$ cz for these two cases the result may exceed the range.
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Now by checking Option B and Option C we can get the ans easily :)