A simple proof is available as well. Pick p coprime to d and let t be such that td=1 mod p. Then, mod p, t times the arithmetic progression looks like a sequence of consecutive integers. Thus its length has to be less than p to avoid one of the terms being a multiple of p, which means the original progression also has to have fewer than p terms. So the collection of primes dividing a set of arbitrarily long arithmetic progressions must be infinite.
It has been noted by GH from MO that the above overlooks some subtleties; the following is
more inspired by Euclid, and should leave no doubt remaining.
Let A be a set of arbitrarily long nonconstant arithmetic progressions. Thus for any
integer m, we can find a member of A and extract from that (wlog) a positive increasing
arithmetic progression of the form a +kd where a and d are coprime and k goes from 0 to
m. Pick a finite set of primes M, set m equal to a large multiple of their product (so at
least twice the product of primes of M) and then choose from A a progression and derive
the progression a +kd with the properties above. I will show the existence of a prime divisor
which is not in M and divides a member of the progression.
Now d may share some factors with the product of M, but as (a,d)=1, none of the factors
of d will divide any of a +kd. To be safe, let us call M' that set of numbers in M which are
coprime to d, and set their product to m'. So (m',d)=1, d is a unit mod m', and we can look
at ta +tkd as k runs from 1 to m which is bigger than m'. As above, td=1 mod m'.
Modulo m', ta+tkd is ta+k, so ta+k is divisible by a factor of m' precisely when a+kd is. As a result,
there are k bigger than m' and at most m such that a+kd is coprime with m'. But a+kd is bigger than
m' and is coprime not just to m' but to the product of all primes in M. So it must have a prime factor
outside of M. So A "contains" more primes than found in any finite set of prime factors.
Clear enough?
Yes, this is unknown; it is even unknown (as GH from MO suspected in a comment) whether $P(p) \ge 3$ always. An equivalent statement to $P(p) \ge 3$ is that there exists an integer $x>0$ such $p+x$ and $p+2x$ are both prime. This is a twin-prime-like problem: nobody has ever proved a statement saying that two fixed linear polynomials $ax+b$ and $cx+d$ are infinitely often simultaneously prime, or even that they must generally be simultaneously prime once. (The Green-Tao theorem converts into a statement about linear polynomials $x,x+d,x+2d,...$ in two variables $x$ and $d$; when we fix $p$ here, we have only one variable.)
On the other hand, the prime $k$-tuples conjecture does imply that $P(p)=p$ for every prime $p$: the corresponding polynomials are $p+x,\dots,p+(p-1)x$, and these polynomials form an admissible set (their product is not identically zero modulo any prime).
Best Answer
It is firmly expected that for every \epsilon > 0 each aritmetic progression with difference q and terms coprime with q will contain a prime <<{\epsilon} q^{1 + \epsilon}. This is a direct consequence of a conjecture of H. L. Montgomery (not the one on pair correlations, another one), which implies both the GRH and the Elliott-Halberstam conjecture. But the upper bound $\ll {\epsilon} q^{1 + \epsilon}$ for the least prime in an arithmetic progression was conjectured by S. Chowla (in his book "The Riemann Hypothesis and Hilbert's tenth problem"), and probably by others independently of him, years before Montgomery made his conjecture, and presumably on the basis of the same kind of heuristic argument that moonface advances. In fact, I think that even an upper bound as strong as qlog(q)^2 has been conjectured, though I won't swear to that. But it is definitely known that qlog(q) won't work. The Montgomery conjecture seems reasonable, because it rests on the assumption that there is square root cancellation in a certain sum with D-characters as coefficients in an explicit formula - this ties in with moonface's comment about the proof that GRH implies L \leq 2 being "lossy". On the other hand, one cannot expect to get q^{1 + \epsilon} out of the Elliott-Halberstam conjecture in any direct way, because that is an averaging kind of statement. You would not expect to get L \leq 2 out of the Bombieri-A. I. Vinogradov theorem either, for that is an averaged version of GRH. The point is that information is needed about every single one of the arithmetic progressions individually.