The statement
$$(p\wedge q),(p\implies r)\models r$$
Means that whenever the formulas $(p\wedge q)$ **and** $(p\implies r)$ are (both simultaneously) true, then $r$ must be true.
So you job is just to write the truth tables of those 3 formulae and verify that in the lines where $[\\![(p\wedge q)]\\!]=[\\![(p\implies r)]\\!]=1$, you also have $[\\![r]\\!]=1$ ($[\\![x]\\!]$ means the truth value of $x$).
Also, in this case, it's pretty intuitive that this is true, the relevant information of the first formula says that $p$ must be true, the second one says that if $p$ is true then $r$ is true, and we have to conclude that $r$ is true, the 'rule' $(p\wedge q )\models p$ is called conjunction elimination and the rule $p,(p\implies r)\models r$ is called modus ponens.