Before solve these type of question remember few things
1. ^ is AND / . 2. v is OR / + 3. A --> B is A'+ B ( if A is 1 and B is 0 then only A--->B is false(0) )
Now, Check Option B. ¬(p->q) --> ¬q
Try to use implication property here
take q' as false(0) and if ¬(p->q) is true(1) then we can say directly its not tautology
so check now, ¬(p->q) --> 0 // q'=0 so, q=1
¬(p->1) --> 0 // p-->1 will be 1 only ,
So, ¬(1) --> 0
0-->0 that will give 1 (True) : TAUTOLOGY.
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Now Check Option A : (¬P^(P->q))->¬q
you can solve it either directly or use above method
here i'm solving direct
so, (P' .(P-->q))' + q'
= P+ (P'+q)' +q'
= P+ P.q' +q'
= P+q' (NOT TAUTOLOGY)
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Similarly you can solve Option C
ANSWER: OPTION B