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I am not very familiar with truth tables but I think that the $\lnot$ should get distributed among both $p$ and $q$ making the problem $\lnot p \implies \lnot q$ which does is not the same as $p\land q$ making the statement false.

I know that $\lnot q \implies \lnot p$ is the contrapositive of $p \implies q$ which is also equivalent to $\lnot p$ or $q$, and if we switch the $p$ and $q$ it will still make it false.

If anyone can confirm my answer or give more of an explanation that would be great as I am very lost!

Thank you to all of the help in advance, it is very appreciated.

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    $\begingroup$ Note that $p \implies q \equiv \neg p \vee q$. So it's negation is $\neg (\neg p \vee q) \equiv p \wedge \neg q$ by De-Morgan's law. $\endgroup$
    – Anurag A
    Commented Jul 15, 2020 at 20:16
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    $\begingroup$ @AnuragA Please post your comment as an answer, as it is correct and exactly what I would have answered. (Though I would also have said explicitly that the proposed equivalence by the OP is false.) $\endgroup$
    – amWhy
    Commented Jul 15, 2020 at 20:24
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    $\begingroup$ To emphasize a part of @amWhy's comment: $\neg$ does not distribute over $\implies$. $\endgroup$
    – Andreas Blass
    Commented Jul 16, 2020 at 17:55

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It is false.

Consider when both $p$ and $q$ are true. Then the RHS is true, whereas, since $p\implies q$ is true, the LHS is false.

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