Solution:

The answer is $\mathbf{( E )}$. Let us set $n=\lfloor p\rfloor$ and $\alpha=\{p\}$. Then the equation can be rewritten as $\{n(n+\alpha)\}=2021 \alpha$, that is,
$$\{n \alpha\}=2021 \alpha$$
Since on the left we have a fractional part, which is $<1$, and moreover $p$ is not an integer, we obtain the conditions $0 \neq \alpha<\frac{1}{2021}$.
From the equation we also derive that there exists a non-negative integer $m$ such that $n \alpha=m+2021 \alpha$. Rearranging, $(n-2021) \alpha=m$.
Now, if $m=0$ we have $n=2021$ which is a solution (one checks that $p=2021+\alpha$ works for every $\alpha<\frac{1}{2021}$), and it is also the smallest one.
If $m \geq 1$, we have $\alpha=\frac{m}{n-2021}<\frac{1}{2021}$, from which $n>2021+2021 m \geq 4042$, and hence $n \geq 4043$. Indeed, one checks that $p=4043+\frac{1}{2022}$ is a solution.