I don't see why you want to solve your problem using the truth table.
- Firstly, here is an intuitive approach that I believe the first thing to do when you have such problems.
$$ ``\mbox{Warning lights is on}" \Longleftrightarrow ``\mbox{Pressure is high}" \wedge \ `` \mbox{Relief valve is clogged }" $$
is equivalent to :
$$ ``\mbox{Warning lights is } \color{#C00}{off} " \Longleftrightarrow `` \mbox{Pressure is } \color{#C00}{not} \mbox{ high}" \vee \ `` \mbox{Relief valve is } \color{#C00}{not} \mbox{ clogged }" \tag{P} $$
Now, we consider the statements $(P)$ to be true and that $``\mbox{the Relief valve is not clogged}"$, that implies that Warning lights is off ( regardless of the pressure).
So the conclusion is obviously invalid (because the pressure can be too hight and the warning still off).
- Now if the table is required by the question, you have to add an arrow in your table that contain the truth values of
$$ \left( (W \Longleftrightarrow R \wedge P) \wedge \lnot R \right) \Longrightarrow \left( W \Longleftrightarrow P \right) $$
For a counterexample you look at the very rows where the premises are all true but the conclusion is false. In particular, look back at the reference columns on the left. For example, row 3 is one of these rows, and in this row $K$ is true, but $J$ and $L$ are false. Typically, that is enough to be the counterexample: you just say: "A counterexample to the validity of this argument would be when $K$ is true, and $J$ and $L$ are false, for then all ther premises are true, but the conclusion is false."
Sometimes, however, they want you to provide a more concrete scenario, i.e. where $J$, $K$, and $L$ have some meaning so that the invalidity of the argument is even more Obvious. So: pick something for $J$, $K$, and $L$ that makes the premises true (in our world), but the conclusion false. OK, how about:
$J$ : Bob is an adult male
$K$ : Bob is unmarried
$L$ : Bob is a bachelor
Clearly, the two premises are now true, since effectively they say that if you are an adult male and you are unmarried, then you are a bachelor. The conclusion, however, says that you are a bachelor as soon as either you are an adult male, or you are unmarried ... and that is not true. I could be an adult male, but not a bachlor, since I am married. Or: I could be unmarried, but still not a bachlor, since I am female.
Anyway, the point is: look at the reference columns for the rows where the premises are true and the conclusion is false to get your counterexample!
Best Answer
Note that the question is not about the formula $$(p\to q)\land q \rightarrow p $$ but about the proposed inference $$ (p\to q), q \vdash p $$ This is important because $\vdash$ is not a connective and shouldn't have a truth value.
In order to use a truth table, as requested by the question, you should have "input" columns $p$ and $q$, and "output" columns $p\to q$ as well as $q$ and $p$:
$$ \begin{array}{cc|ccc} p & q & p\to q & q & p \\ \hline F & F & T & F & F \\ T & F & F & F & T \\ F & T & \cdots \\ T & T & & \cdots \end{array} $$
If you can find a row where the $p \to q$ and $q$ outputs are both T and the $p$ output is $F$, then your truth table shows that the inference $(p\to q), q\vdash p$ is not valid.