For the minimum, notice that by induction $f(p_i)-f(p_j)\geq i-j$ for every $i-j$ ($p_1<p_2<..$ is the sequence of the of primes (corresponding set is $\mathbb{P}$)). So $f(2020)=f(2)f(5)f(101)\geq f(2)((f(2)+2)(f(2)+25)=\geq 2*4*27=216$. For the maximum let $K=f(\mathbb{P})$ then we want to make sure that the elements of K are as big as possible to maximize $f(2020)$. So of course since we already have to obtain all prime powers , it wouldn't make sense for a numbers with more than two distinct primes to be in K. Therefore we can restrict to the powers of prime $p$ which are in K and let's call them $p^{a_1}<p^{a_2}<..$. since we have to obtain $p$, we must have $a_1=1$. Also we have $a_n\leq a_{n-1}+...+a_1+1$ because otherwise we couldn't obtain $p^{a_1+..+a_{n-1}+1}$. So in the best case equality holds, which yields the sequence $a_n=2^{n-1}$, by the identity $2^{n+1}-1=2^n+...+2+1$. So $K=\{p^{2^k}| p\in \mathbb{P}, k\geq 0\}=\{k_1,k_2,..\}$, with $k_1<k_2<...$. Therefore $f(2020)\leq k_1*k_3*k_{26}=584$. So $216\leq f(2020)\leq 584$