What we would like to prove is a conjunction, so it suffices to prove each conjunct separately and then glue them together at the end. This problem would probably be easier and more intuitive using proof by contradiction, but after talking with the asker, I will provide a direct proof.
$$\begin{array}{lr}
1. & (P \rightarrow Q)\wedge(Q \rightarrow R) & \text{Premise} \\
2. & P \rightarrow Q &\text{Simplification, 1}\\
3. & Q \rightarrow R & \text{Simplification, 1}\\
4. & \neg{P} \vee Q & \text{Conditional Law, 2}\\
5. & \neg{Q} \vee R & \text{Conditional Law, 3}\\
6. & Q \vee \neg{Q} & \text{Tautology} \\
7. & \neg{P} \vee R & \text{Constructive Dilemma, 4,5,6}\\
8. & P \rightarrow R & \text{Conditional Law, 7}\\
9. & (P \rightarrow Q) \vee (Q \rightarrow R) &\text{Addition, 2}\\
10. & (Q \rightarrow P) \vee (Q \rightarrow R) &\text{Addition, 3}\\
11. & (P \rightarrow Q) \vee (R \rightarrow Q) &\text{Addition, 2}\\
12. & Q \vee \neg{Q} & \text{Tautology}\\
13. & (Q \vee \neg{Q}) \vee (P \vee \neg{R}) & \text{Addidition, 12}\\
14. & (\neg{Q} \vee P) \vee (\neg{R} \vee Q) & \text{Associative Law, 13}\\
15. & (Q \rightarrow P) \vee (R \rightarrow Q) & \text{Conditional Law, 14}\\
16. & \big((P \rightarrow Q) \vee (Q \rightarrow R)\big)\wedge \big((Q \rightarrow P) \vee (Q \rightarrow R)\big) & \text{Conjunction, 9,10}\\
17. & \big((P \rightarrow Q) \vee (R \rightarrow Q)\big)\wedge \big((Q \rightarrow P) \vee (R \rightarrow Q)\big) & \text{Conjunction, 11,15}\\
18. & \big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (Q \rightarrow R) & \text{Distributive Law, 16}\\
19. & \big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (R \rightarrow Q) & \text{Distributive Law, 17}\\
20. & \Big(\big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (Q \rightarrow R)\Big) \wedge & \\
&\Big(\big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee (R \rightarrow Q)\Big) & \text{Conjunction, 18,19}\\
21. & \big((P \rightarrow Q) \wedge (Q \rightarrow P)\big)\vee \big((Q \rightarrow R) \wedge (R \rightarrow Q) \big) & \text{Distributive Law, 20}\\
22. & (P \equiv Q) \vee (Q \equiv R) & \text{Definition of Biconditional, 21}\\
\therefore & (P \rightarrow R)\wedge \big((P \equiv Q) \vee (Q \equiv R)\big) & \text{Conjunction, 8,22}
\end{array}$$
As desired.
\begin{array}{rl} & & \neg(P\land Q)\to(\neg P\lor(\neg P\lor Q)) \iff\\ & \iff & (P\land Q)\lor(\neg P\lor Q) \iff & \quad & \text{Material Implication}\\ & \iff & (\neg P\lor Q\lor P) \land(\neg P\lor Q\lor Q) & \quad & \text{Distributive Law}\\ \end{array}
As you can verify, $\neg P \vee Q \vee P$ is a tautology.
$$\neg P \vee Q \vee P \iff \neg P \vee P \vee Q \iff \top \vee Q \iff \top$$
Hence, we have the following.
\begin{array}{rl} & & \top \wedge (\neg P \vee Q \vee Q) \iff\\ & \iff & \neg P \vee Q \vee Q \iff\\ & \iff & \neg P \vee Q & \square\\ \end{array}
Best Answer
The given statement has one more opening parenthesis than closing parenthesis .... but I'll go with:
$(p\lor (r\lor q))\land \neg(\neg q\land\neg r)$
OK, like you said, DeMorgan seems like a good first step:
$(p\lor (r\lor q))\land (\neg\neg q\lor \neg \neg r)$
Two double negations gives:
$(p\lor (r\lor q))\land(q\lor r)$
By Commutation:
$(p\lor ( q \lor r))\land(q \lor r)$
Absorption:
$q \lor r$
Unfortunately, some textbooks do not give you Absorption. If not, you can do:
$(p \lor (q \lor r)) \land (q \lor r)$
Identity:
$(p \lor (q \lor r)) \land (\bot \lor (q \lor r))$
Distribution:
$(p \land \bot) \lor (q \lor r)$
Annihilation:
$\bot \lor (q \lor r)$
Identity:
$q \lor r$