Let $k > 1$ be any divisor of the product $a_1 \dots a_n$. Then $k$ has a prime divisor $p$, which is also a divisor of the product $a_1 \dots a_n$. As $p$ is a prime, there exists $i$, such that $p$ is a divisor of $a_i$. As all the divisors of $a_i$ greater than $1$ have the property, $p$ also has this property. By the premise, all the multiples of $p$ have the property, so $k$ has the property.

Let $k > 1$ be any divisor of the product $a_1 \dots a_n$. If $k$ were relatively prime to all $a_i$, it would be relatively prime to the product $a_1 \dots a_n$, but $\text{gcd}(k, a_1 \dots a_n) = k > 1$. Hence $\text{gcd}(k, a_i) > 1$ for some $a_i$. As a divisor of $a_i$, the number $\text{gcd}(k, a_i)$ has the property studied by professor $P$. As a multiple of $\text{gcd}(k, a_i)$, also $k$ has the same property.