= Solution
Let $G$ be the <Galois group of a polynomial> of the irreducible quartic. Irreducibility makes $G$ a transitive subgroup of $S_4$. The three roots of the given <cubic resolvent of a quartic> correspond to the three partitions of the four quartic roots into two unordered pairs, and the kernel of the $S_4$-action on these partitions is the <Klein four-group> $V_4$. If the resolvent has Galois group $S_3$, the image of $G$ in this action is all of $S_3$. Among the <transitive subgroups of the symmetric group on four points>, only $S_4$ has this image. Hence
$$
\boxed{\ \operatorname{Gal}(f/\mathbb Q)\cong S_4.\ }
$$
Now put $f_p(t)=t^4+pt+p$. The <Eisenstein criterion> at $p$ proves that $f_p$ is an <irreducible polynomial>, and its cubic resolvent is
$$
g_p(t)=t^3-4pt-p^2.
$$
Its <discriminant of a depressed cubic> is
$$
\Delta_p=256p^3-27p^4=p^3(256-27p).
$$
For odd $p$, the $p$-adic valuation of $\Delta_p$ is three because $p\nmid256$; for $p=2$, $\Delta_2=2^4\cdot101$. Thus $\Delta_p$ is never a square in $\mathbb Q$.
By the <rational root theorem>, a reducible $g_p$ has an integer root among $\pm1,\pm p,\pm p^2$. Reduction modulo $p$ shows that the root is divisible by $p$; writing it as $pk$, with $k\in\{\pm1,\pm p\}$, gives
$$
pk^3-4k-1=0.
$$
The only possibilities are $k=-1,p=3$ and $k=1,p=5$, corresponding to roots $-3$ and $5$. For every other prime, $g_p$ is irreducible with nonsquare discriminant, so its Galois group is $S_3$ and the criterion above gives the quartic group $S_4$. For $p=3$ or $5$, the reducible resolvent makes the quartic group a proper subgroup. Therefore
$$
\boxed{\ p=3\ \text{or}\ p=5.\ }
$$
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