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.
You should start supposing the contrary of the conclusion, that is,
$$\neg(\neg p \vee (\neg q\vee r)))$$
Now, apply twice the De Morgan's law for $\neg(X\vee Y)$:
$$p\wedge q \wedge\neg r$$
Eliminate $\wedge$ to obtain $\neg r$. Now, use the premise and eliminate $\vee$:
$$\neg(\neg p\vee q)$$
Apply again De Morgan's law:
$$p\wedge \neg q$$
Now, eleminate $\wedge$ in two previous formulae to get $q$ and $\neg q$. Insert $\wedge$ and you have the contradiction.
Best Answer
Using distibution we have: $$ p \lor ( \neg p \land q ) \equiv \big(p\lor (\neg p)\big)\land \big(p\lor q\big) \equiv 1 \land \big(p\lor q\big) \equiv p\lor q $$ Then De Morgan gives answer