Using that equivalence $A\to B\equiv \
eg A \lor B$, and the fact(s) that $\land, \lor$ are commutative, the hypothesis is equivalent to $$ (\
eg r \to \
eg p) \land (\
eg p \to q),\tag{Hyp.} $$ (For example, ($\
eg r \to \
eg p) \equiv (\
eg p \lor r)$.) This is the premise of a syllogism. From (Hyp.), you can infer $\
eg r \to q$, or equivalently, after eliminating double negation and using commutativity or $\lor$, $$ q \lor r. \tag{*} $$ (Every instance of $[(A\to B)\land(B\to C)]\to (A\to C)$ is a tautology, a theorem, is valid; so from that together with (Hyp.), (*) follows by Modus Ponens.)
This shows that $$ (p \lor q) \land (\
eg p \lor r) \vdash (q \lor r), $$ so the result follows from the Deduction Theorem.