These are fairly standard terms, but for the sake of completeness: An ultrafilter $\mathcal{U}$ on $\omega$ is a *p-point* if whenever $(A_n)_{n<\omega}$ is a partition of $\omega$ such that $A_n \notin \mathcal{U}$ for all $n$, there is an $X \in \mathcal{U}$ such that $X \cap A_n$ is finite for all $n$. $\mathcal{U}$ is *Ramsey* if the same holds but with $|X\cap A_n| = 1$ for all $n$. Clearly Ramsey ultrafilters are p-point ultrafilters. The name Ramsey comes from the fact that if $\mathcal{U}$ is a Ramsey ultrafilter, then any graph on $\omega$ has a homogeneous set in $\mathcal{U}$.

Fix an edge relation $E$ on $\omega$ making it into a copy of the random graph. (For concreteness, we could say that $nEm$ if and only if the $\min(n,m)$th binary bit of $\max(n,m)$ is $1$.) A remarkable property of the random graph is that for any finite partition $X_0,X_1,\dots,X_{n-1}$ of $\omega$, there is an $i<n$ such that $(X_i,E)$ is isomorphic to $(\omega,E)$. (See Proposition 3 here.) This implies the following.

Proposition.There is an ultrafilter $\mathcal{U}$ on $\omega$ with the property that for any $X \in \mathcal{U}$, there is a $Y \subseteq X$ such that $(Y,E)$ is isomorphic to $(\omega,E)$.

*Proof.* For each finite partition $P$ of $\omega$, let $F_P$ be the (clopen) set of ultrafilters $\mathcal{U}$ in $\beta \omega \setminus \omega$ satisfying that for the unique $X$ in $\mathcal{U} \cap P$, there is a $Y \subseteq X$ such that $(Y,E)$ is isomorphic to $(\omega,E)$. By the above fact, each $F_P$ is non-empty. The family $\\{F_P:P\text{ a finite partition of }\omega\\}$ has the finite intersection property. To see this, note that if $P_0,P_1,\dots,P_{n-1}$ are finite partitions of $\omega$, then for any mutual refinement $Q$ of the $P_i$'s, $F_Q \subseteq \bigcap_{i<n} F_{P_i}$. Therefore, by compactness, there is an ultrafilter $\mathcal{U}$ such that for every $P$, $\mathcal{U} \in F_P$. For any set $X \in \mathcal{U}$, we have that $\mathcal{U} \in F_{ \\{X,\omega\setminus X\\} }$, so $X$ has the required property. $\square$

Clearly, any such ultrafilter cannot be Ramsey, since an $E$-homogeneous set will contain no subset isomorphic to the random graph. Call such an ultrafilter *aggressively non-Ramsey*.

Question.Is it consistent with $\mathsf{ZFC}$ that there is an aggressively non-Ramsey p-point?