Here's the picture of the question:
How does it go from
p v ~q to
~p -> ~q?
$$p\lor \lnot q$$ $$ \equiv \lnot\lnot p \lor \lnot q$$ $$\equiv \lnot p \rightarrow \lnot q $$
This is simply applying the rule that $a \rightarrow b \;\equiv \; \lnot a \lor b$,
In this case, $a$ happens to be $\lnot p$, and $b$ happens to be $\lnot q$.
To double-check, and convince yourself, truth-tables come in handy:
One last note:
The right-hand side is called the contrapositve of the left-hand side of the equivalence; they are equivalent expressions.
Knowing this, one can conclude, directly, that the converse of $p \rightarrow q$ is equivalent to the inverse of $p \rightarrow q$: $(q\rightarrow p) \equiv (\lnot p \rightarrow \lnot q)$.
An easy way to remember what $p \Rightarrow q$ means is this: if the implication is true, either the conclusion is true, or the premise is false. That is, $$p\Rightarrow q \equiv \lnot p \lor q$$ So $p \lor \lnot q \equiv \lnot \lnot p \lor \lnot q \equiv \lnot p \Rightarrow \lnot q$
We'd half to know more about this particular text to know how the author(s) did this. That said, following for example Lukasiewicz's introductory text on mathematical logic, you can define (p$\lor$q) as ($\lnot$p$\implies$ q), since they are logically equivalent... (p$\lor$q)$\equiv$($\lnot$p$\implies$ q).
Then from that equivalence just uniformly substitute $\lnot$q for q, and you have (p$\lor$$\lnot$q)$\equiv$($\lnot$p$\implies$$\lnot$q).