Here is a concrete example:
Suppose we have the following truth-table ($P * Q$ is some arbitrary formula involving atomic propositions $P$ and $Q$):
\begin{array}{cc|c}
P & Q & P * Q\\
\hline
T & T & T\\
T & F & T\\
F & T & F\\
F & F & T\\
\end{array}
Note that the formula is true in rows 1,2, and 4. In row 1, $P$ and $Q$ are both true, so we generate the corresponding expression $P \land Q$. In row 2, $P$ is true and $Q$ is false, so we generate the term $P \land \neg Q$. Row 4 corresponds to $\neg P \land \neg Q$.
If we now disjunct all these terms together, we get:
$$(P \land Q) \lor (P \land \neg Q) \lor (\neg P \land \neg Q)$$
Now see what happens if we put this expression back on a truth-table:
\begin{array}{cc|c|c|c|c}
P & Q & P \land Q & P \land \neg Q & \neg P \land \neg Q & (P \land Q) \lor (P \land \neg Q) \lor (\neg P \land \neg Q)\\
\hline
T & T & T & F & F & T\\
T & F & F & T & F & T\\
F & T & F & F & F & F\\
F & F & F & F & T & T\\
\end{array}
Note how each of the disjuncts is true in exactly the one row that it was generated from, and thus how the disjunction is true in exactly those rows where the original expression $P * Q$ is true. Thus, we can describe $P * Q$ with the formula $(P \land Q) \lor (P \land \neg Q) \lor (\neg P \land \neg Q)$. I think you will understand why this process will always work, no matter what the function that needs to be described looks like, and no matter how many atomic propositions are involved.
One last thing: if there is no row where the formula is true, then with this process you would not get any terms. However, this also means that the original formula is a contradiction, and we can always describe a contradiction with $P \land \neg P$ .. which is in DNF.
Okay, let me see if I can walk you through this. First of all, I(P) just means that I is a function that tells you if the literal propositional variable P is true or false, and returns 1 or 0 accordingly. (Maybe the I stands for interpretation? Check your text.)
DNF is the more straightforward one intuitively. It's describing the truth table where you want to OR together a bunch of conjunctions of propositional variables (or their negations). It's exactly how you would describe the truth table to someone else. In the table you listed, you would say that the final statement is true iff P is true and Q is true and R is false OR P is false and Q is true and R is false. Symbolically, you would write that as $(P\wedge Q\wedge\neg R)\vee(\neg P\wedge Q\wedge\neg R) $.
CNF is the opposite, where you're ANDing together a bunch of "rules" instead of ORing together a bunch of "cases". The general way of making a CNF expression for a formula $A$ is to make the DNF of $\neg A$ and then use the De Morgan laws to propogate that negation down to the literals, which switches all of the conjunctions to disjunctions and vice-versa.
It's been a while, so I might be missing a detail or two, but hopefully that gives you some intuition about how and why you'd want to do this.
Best Answer
This example is for explanation of what a Disjunctive Normal form is, On purpose i don't simplify the resulting formula, The question is about homework and it should stay homework.
suppose you have the truth table:
and you want to transform that into a formula in the Disjunctive Normal form.
First add to the true values the formula that makes that row valid
(sorry had some problems with the layout)
and then or them together
$ (P \land Q ) \lor (P \land \neg Q ) \lor (\neg P \land \neg Q ) $
And this formula is in Disjunctive Normal form.
notice this formula can be simplified but that is out of scope for this answer, see
http://en.wikipedia.org/wiki/Disjunctive_normal_form
and http://en.wikipedia.org/wiki/Karnaugh_map for that.
Good luck