Not sure whether this is of help after such a long time, but anyway:
The answer is yes, and this even works over some number field in which all the primes in $P$ are completely split. Namely, denote by $K^P$ the maximal algebraic extension of $\mathbb{Q}$ in which all $p\in P$ split completely.
As noted by Pop on p.3 here: https://www.math.upenn.edu/~pop/Research/files-Res/LF_6Oct2013.pdf ,
the field $K^P$ is a "large" field, meaning that every irreducible curve over $K^P$ with a smooth $K^P$-point has infinitely many such points (I suppose for this particular field, the "large" property actually follows from some variant of Krasner's lemma). The regular inverse Galois problem is known to have a positive answer over all large fields (due to e.g. Pop, Harbater, Jarden etc.), so there is a regular $G$-Galois extension of $\mathbb{P}^1(K^P)$. But this extension must of course be defined over some number field $F\subset K^P$, giving the answer.
Thank you for calling this problem to my attention.
I computed $K$ en route to AWS (though this year's
topics
are a rather different flavor of number theory...).
After some simplification (gp's $\rm polredabs$), it turns out that
the field $K$ is generated by a root of
$$
f(x) = x^{17} - 2x^{16} + 8x^{13} + 16x^{12} - 16x^{11} + 64x^9 - 32x^8 - 80x^7
$$
$$
\qquad\qquad\qquad {} + 32x^6 + 40x^5 + 80x^4 + 16x^3 - 128x^2 - 2x + 68.
$$
gp quickly confirms:
p = Pol([1,-2,0,0,8,16,-16,0,64,-32,-80,32,40,80,16,-128,-2,68])
poldegree(p)
F = nfinit(p);
factor(F.disc)
returns
[2 79]
So the field has discriminant $2^{79}$.
Let $F = {\bf Q}(\zeta_{64}^{\phantom{0}} - \zeta_{64}^{-1})$,
with $F/\bf Q$ cyclic of degree $16$ of $\bf Q$.
It was known that $K$ is contained in
an unramified cyclic extension $L/F$ of degree $17$.
Hence $L(\zeta_{17})$ is a Kummer extension of $F_{17} := F(\zeta_{17})$.
Now $F_{17}$ is a $({\bf Z} / 16 {\bf Z})^2$ extension of $\bf Q$.
Fortunately it was not necessary to work in the unit group
of this degree $256$ field: the Kummer extension is
$F_{17}(u^{1/17})$ with $u \in F_{17}^* / (F_{17}^*)^{17}$
in an eigenspace for the action of ${\rm Gal}(F_{17}/{\bf Q})$,
which makes it fixed by some index-$16$ subgroup of this Galois group.
Thus $u$ could be found in one of the cyclic degree-$16$ extension of $\bf Q$
intermediate between $\bf Q$ and $F_{17}$, and eventually a generator of $K$
turned up whose minimal polynomial, though large, was tractable enough for
gp's number-field routines to do the rest.
Best Answer
This is an elaboration of Chris Wuthrich's comment. Let $p$ be unramified (i.e. $p$ does not divide the discriminant of the Galois extension $K / \mathbb{Q}$), and let $\mathfrak{P}$ be a prime in $\mathcal{O}_K$ over $p$ with decomposition group $D_\mathfrak{P}\leq G$. The primes in $\mathcal{O}_K$ over $p$ correspond bijectively to the left cosets in $G/D_\mathfrak{P}$, and the action of $G$ on them corresponds to the left action of $G$ on $G/D_\mathfrak{P}$. Now $D_\mathfrak{P}$ is the cyclic group generated by $\mathrm{Frob}_\mathfrak{P}$, and changing $\mathfrak{P}$ results in conjugating $\mathrm{Frob}_\mathfrak{P}$. Hence the question boils down to how often the conjugacy class of $\mathrm{Frob}_\mathfrak{P}$ equals a given conjugacy class $C\subset G$. By the Chebotarev density theorem, the density of such primes $p$ equals $|C|/|G|$.