An one-sentence proof of P ⊆ NP
Since L is in P, you can answer the word problem in polynomial time. To show that L is in NP as well, we need to provide a polynomial checking relation $R$ such that

$$ w\in L \Leftrightarrow \exists y.(|y|\le |w^k| \text{ and } R(w,y))$$

Now Prof. Cook says to take a very simple $R$. For every $w$ in $L$, no matter what $y\in \Sigma^*$ you take, $R(w,y)$ is true and for every $w$ not in $L$, $R(w,y)$ is false, regardless of the $y$. This is a polynomial time relation, since we can decide whether $w\in L$ or not in polynomial time (since $L \in P$), without looking at $y$ at all. And as any $y$ works, there are also some $y$ that are short enough to satisfy the length restriction in the above definition.