Basically, I would agree with @boboquack that a truth table is the way to go. What you want to say goes beyond what you can represent reasonably in propositional logic. You have people and you have statements. The people make statements about statements. You really want to say something about the relationship of the people to the statements you make and for this you need a more complex notation. You want something like this:
$$\forall (x) (KNIGHT(x) \rightarrow (\forall (y) (SAYS(x, y) \rightarrow y))) $$
That is, anyone who is a knight makes only true statements or "For ALL $x$, IF $x$ is a KNIGHT THEN for ALL $y$, IF $x$ SAYS $y$ THEN $y$ is true." So, $SAYS(x,y)$ means that the "$SAYS$" relationship holds between $x$ and $y$.
Similarly,
$$\forall (x) (\lnot KNIGHT(x) \rightarrow (\forall (y) (SAYS(x, y) \rightarrow \lnot y))) $$
If $x$ is not a knight, and $x$ says $y$ then $y$ is not true.
The first statement made by $a$ becomes:
$$ SAYS(a, \lnot KNIGHT(a) \land \lnot KNIGHT(b) \land \lnot KNIGHT(c)) $$
That is, $a$ says that $a$ is not a knight and $b$ is not a knight and $c$ is not a knight. You can show that assuming the truth of this statement leads to a contradiction. To be thorough, you need to translate all the information into logical notation: $a$, $b$, and $c$ are persons and everyone who is a person must be either a knight or a knave. I leave this for you to work out.
The second statement is also a bit tricky. I would translate part of it as:
$$ SAYS(b, KNIGHT(a) \rightarrow (\lnot KNIGHT(b) \land \lnot KNIGHT(c)) $$
If $a$ is a knight then neither $b$ nor $c$ is a knight. He is also making similar statements about the knighthood of $b$ and $c$. Put this all together and you will (eventually) arrive at the desired conclusion.
A truth table is way easier.
If you want to stick to propositional logic you could write $a$'s first statement as:
$$\lnot PA \land \lnot PB \land \lnot PC$$
where $PA$ represents the statement "$a$ is a truth teller". Similarly for $PB$ and $PC$. You then have to know that $\lnot PA$ implies that this statement itself is false which implies that $PB$ or $PC$ is true... and so on.