$P$ is the set of prime numbers and I may assume that there are infinitely many prime numbers. My first thought was using an infinite grid but the question is only $2$ marks and I think I have spotted an easier method using the assumption stated in the question. I know that $P \subset N$, and we know that $N$ is countably infinite since there exists a bijection $f: N \to N$.
By the lemma for any sets $X$ and $Y$ with $X \subset Y$, we have $|X| \leq |Y|$, I now have $|P| \leq |N|$, but since $P$ is an infinite set, then we must have $|P| \geq |N|$. So in all, $|P|=|N|$, which means $P$ is countably infinite since it has the same cardinality as $N$.
The right hand side shows how the elements of this grid may be labelled with elements of $N$, by proceeding down each diagonal in turn from top left to bottom right. This creates a bijection from $N$ to $P \times P$, namely the function $f : P \to P \times P$ in which each $n \in N$ in the right hand grid maps to the pair $(a, b)$ in the corresponding position in the left hand grid. $f$ is well-defined since every natural number appears once in the right hand grid and so is mapped to exactly one pair in the left-hand grid; $f$ is injective and surjective since each pair appears precisely once in the left-hand grid so is mapped to by exactly one element of $N$. The existence of this bijection proves that $P \times P$ is countably infinite