Recently Malliaris and Shelah proved that $\mathfrak{p}=\mathfrak{t}$ (see: http://math.uchicago.edu/~mem/Malliaris-Shelah-CST.pdf). Their results are far more general, however for the specify context of this question what they showed is that if $G$ is an ultrafilter on $\omega$, then the ultra product $\omega^\omega/G$ is a linear order with no $(\kappa,\lambda)$-gaps with $\kappa,\lambda<\mathfrak{t}$.

They can obtain the desired equality combining it with a result of Shelah on the existence of a certain type of small gap (i.e. with both sequences in the gap of size smaller than $\mathfrak{t}$) in structures of the form $\omega^\omega/G$ in the case $\mathfrak{p}<\mathfrak{t}$.

Recall that

$\mathfrak{t}$ is the least size of a decreasing chain in $P(\omega)/Fin$ without a positive lower bound in the boolean algebra $P(\omega)/Fin$.

$(f_\alpha:\alpha<\kappa,g_\beta:\beta<\lambda)$ is a $\kappa,\lambda$ pregap in a partial order $(X,<^*)$ if $f_\alpha<^*f_\beta<^*g_\eta<^*g_\gamma$ for all $\alpha<\beta<\kappa$ and $\gamma<\eta<\lambda$.

A pregap $(f_\alpha:\alpha<\kappa,g_\beta:\beta<\lambda)$ on $(X,<^*)$ is a gap if for no $f\in X$ we have that $f_\alpha<^*f<^*g_\gamma$ for all $\alpha<\kappa$ and $\gamma<\lambda$.

A gap $(f_\alpha:\alpha<\omega_1,g_\beta:\beta<\omega_1)$ on the partial order $(\omega^\omega,<^*)$ where $<^*$ is eventual domination is an Hausdorff gap if for all $\beta<\omega_1$ and $n<\omega$

$\{\alpha<\beta:\forall m>n \: f_\alpha(m)<g_\beta(m)\}$ is finite.

Malliaris and Shelah's result has also the following consequence on Hausdorff gaps of which I was not aware and which I ask if it had a proof prior to their result (may be even without extra set theoretic assumptions).

Assume $\mathfrak{t}>\omega_1$, $(f_\alpha:\alpha<\omega_1,g_\beta:\beta<\omega_1)$ is an Hausdorff gap.

Let $X\subset P(\omega)$ be the set of $A$ infinite subsets of $\omega$ such that $(f_\alpha\restriction A:\alpha<\omega_1,g_\beta\restriction A:\beta<\omega_1)$ is no longer a gap in the partial order $(\omega^A,<^*)$.

Then $X$ is open dense in the Ellentuck topology.