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.
We verify b) and c) (De Morgan's laws) using a) (double-negation law).
a) $\lnot (\lnot P) \leftrightarrow P$.
b) - Start with the left-hand side and put $\lnot \lnot P$ in place of $P$ and $\lnot \lnot Q$ in place of $Q$ (i.e., use double-negation a)) :
$\lnot (P \lor Q) \leftrightarrow \lnot (\lnot \lnot P \lor \lnot \lnot Q)$
then use c) to transform the content of right-hand side parentheses into : $\lnot (\lnot P \land \lnot Q)$ [ rewrite it as : $\lnot [\lnot (\lnot P) \lor \lnot (\lnot Q) ]$ ; now it is of the "form" : $\lnot [\lnot P_1 \lor \lnot Q_1]$; then you must replace $\lnot P_1 \lor \lnot Q_1$ with $\lnot (P_1 \land Q_1)$, by c), that is really : $\lnot (\lnot P \land \lnot Q)$]. In this way you will get :
$\lnot (P \lor Q) \leftrightarrow \lnot (\lnot \lnot P \lor \lnot \lnot Q) \leftrightarrow \lnot \lnot (\lnot P \land \lnot Q)$
then apply again double-negation to the right-hand side ("cancelling" $\lnot \lnot$) and you will have :
$\lnot (P \lor Q) \leftrightarrow (\lnot P \land \lnot Q)$.
c) - Start with the left-hand side and put $\lnot \lnot P$ in place of $P$ and $\lnot \lnot Q$ in place of $Q$ (i.e., use double-negation a)) :
$\lnot (P \land Q) \leftrightarrow \lnot (\lnot \lnot P \land \lnot \lnot Q)$
then use b) to transform the content of right-hand side parentheses into : $\lnot (\lnot P \lor \lnot Q)$ getting :
$\lnot (P \land Q) \leftrightarrow \lnot (\lnot \lnot P \land \lnot \lnot Q) \leftrightarrow \lnot \lnot (\lnot P \lor \lnot Q)$
then apply again double-negation and it's done.
Best Answer
With the laws that you provide you will not ba able to prove their equivalence. You need an equivalence involving implications. here is the one that is typically used:
Implication: $p \rightarrow q \equiv \neg p \lor q$
Use it as follows:
$(p \land q) \rightarrow r \equiv$ (implication)
$\neg (p \land q) \lor r \equiv$ (deMorgan)
$(\neg p \lor \neg q) \lor r \equiv$ (Idempotence)
$(\neg p \lor \neg q) \lor (r \lor r) \equiv$ (Association)
$\neg p \lor ( \neg q \lor (r \lor r)) \equiv$ (Association)
$\neg p \lor ((\neg q \lor r) \lor r) \equiv$ (commutation)
$\neg p \lor (r \lor (\neg q \lor r)) \equiv$ (Association)
$(\neg p \lor r) \lor (\neg q \lor r) \equiv$ (implication)
$(p \rightarrow r) \lor (q \rightarrow r)$