Statement a
let $G(x)$ : Gora gets the job,
$P(x)$ : Gora gets promoted,
$H(x)$: Gora is happy,
$W(x)$: Gora works Hard.
If Gora gets the job and works hard, then he will be promoted. if Gora gets promotion, then he will be happy. He will not be happy, therefore, either he will not get the job or he will not work hard.
$\equiv$ If Gora gets the job and works hard, then he will be promoted.
if Gora gets promotion, then he will be happy.
He will not be happy,
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$\therefore$ either he will not get the job or he will not work hard.
$\equiv$ $ [G(x) \Lambda W(x)] \rightarrow P(x) $
$P(x)\rightarrow H(x)$
$\sim H(x)$
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$\therefore$ $\sim G(x)\ V \sim H(x)$
$\because$ It is given that $\sim H(x)$ is True so $\sim G(x)\ V \sim H(x)$ will always be True
$\therefore$ Statement $a$ is valid.
Statement b
let $Pu$ : Puneet is guilty.
$Pa$ : Pankaj is telling truth.
Either Puneet is not guilty or Pankaj is telling the truth. Pankaj is not telling the truth, therefore, Puneet is not guilty."
$\equiv$ Either Puneet is not guilty or Pankaj is telling the truth.
Pankaj is not telling the truth
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$\therefore$ Puneet is not guilty.
$\equiv$ $\sim Pu\ V Pa$
$\sim Pa$
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$\therefore\ \sim Pu$
This is the case of disjunctive syllogism
$\therefore$ Statement $b$ is valid.
Statement c
Let $p$ : $n>1$
$q$: $n^{2}>1$
where $n$ is a real number.
If $n$ is a real number such that $n>1$, then $n^{2}>1$. Suppose that $n^{2}>1$, then $n>1$
$\equiv$ If $n$ is a real number such that $n>1$, then $n^{2}>1$
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Suppose that $n^{2}>1$, then $n>1$
$\equiv$ $ p\rightarrow q$
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$\therefore$ $ q\rightarrow p$
We know that if $ p\rightarrow q$ is true then $ q\rightarrow p$ need not be true.
$\therefore$ Statement $c$ is not valid.
Hence $a$ and $b$ are valid statements and $c$ statement is not valid so no option is correct .