No. There are two cases. Firstly, suppose that one of the non-constant coefficients of $P$ is irrational. Then, by the Weyl equidistribution theorem, $\lfloor P(n) \rfloor$ is equidistributed mod $W$ for any modulus $W$, which already limits the natural density of the prime-producing $n$ to be at most $\phi(W)/W$ for any $W$, which implies zero density by taking $W$ to be a product of all the primes less than a large threshold $w$.

If the non-constant coefficients are all rational, then by passing to a suitable arithmetic progression one can make them all integer, at which one one may as well make the constant coefficient integer as well. Then one can sieve using the Chebotarev density theorem (or Landau prime ideal theorem) as in David's answer.

No. There are two cases. Firstly, suppose that one of the non-constant coefficients of $P$ is irrational. Then, by the Weyl equidistribution theorem, $\lfloor P(n) \rfloor$ is equidistributed mod $W$ for any modulus $W$, which already limits the natural density of the prime-producing $n$ to be at most $\phi(W)/W$ for any $W$, which implies zero density by taking $W$ to be a product of all the primes less than a large threshold $w$.

If the non-constant coefficients are all rational, then by passing to a suitable arithmetic progression one can make them all integer, at which point one may as well make the constant coefficient integer as well. Then one can sieve using the Chebotarev density theorem (or Landau prime ideal theorem) as in David's answer. (One should probably get an upper bound of $O(x/\log x)$ for the number of $n \leq x$ with $P(n)$ prime by this method, where the implied constants depend on the coefficients of $P$ of course.)