In fact, the answer to this question as written is easily seen to be no, since there are no solutions to $2^k\equiv -1\pmod{7}$, and hence this fails with $p=2$ and $q=7$. It is possible that Erdős meant to exclude such obstructions, by amending this to 'odd primes' or 'all sufficiently large primes' or such.
Even with such amendments, this problem is false is a strong sense: Alan Tong has provided the following elegant elementary proof that, for any given prime $p$, there are infinitely many primes $q$ such that this statement is false: let $m$ be the product of all primes $\leq p$, and choose a prime $q$ congruent to $-1$ modulo $4m$. If $p$ is the greatest prime divisor of $n$ then, using quadratic reciprocity, every prime divisor of $n$ is a quadratic residue modulo $q$, and hence $n$ is a quadratic residue modulo $q$. On the other hand, since $q\equiv 3\pmod{4}$ we know that $-1$ is not a quadratic residue modulo $q$, and hence $n\not\equiv -1\pmod{q}$, so it is impossible for $q\mid n+1$.
Tong asks whether, for any given odd prime $q$, there are infinitely many primes $p$ such that there is no integer $n$ with $P(n)=p$ and $P(n+1)=q$.
Sampaio independently observed that the answer to Erdős' original problem is no if one of the primes can be $2$ - for example this is false with $p=19$ and $q=2$, since if $n+1=2^k$ and $19\mid n$ then (since $2$ is a primitive root modulo $19$) we must have $18\mid k$, and hence $73\mid 2^{18}-1\mid n$. A similar argument works with $19$ replaced by any prime $p>13$ for which $2$ is a primitive root, using a result of Rotkiewicz [Ro64b] that for every prime $p>13$ there is a prime $q>p$ which divides $2^{p-1}-1$.
Problem 6 in the 12th Romanian Master of Mathematics Competitions in 2020 was to prove that there exist infinitely many odd primes $p$ such that, for every $n$, $P(n)P(n+1)\neq 2p$.
