Since we care about probability measures, a sufficient condition for compactness is that the supports of the respective probability measures are compact. Here is the reasoning:

Let $X$ be a compact subset of $\mathbb{R}^d$. Denote by $C_0^\ast(\mathbb{R}^d)$ the space of all continuous functions on $\mathbb{R}^d$ that vanish at infinity. Then the Banach-Alaoglu theorem implies that the unit ball of $C^\ast_0(\mathbb{R}^d)$ is compact in the weak$^\ast$-topology. This implies that the set of probability measures is compact in the weak$^\ast$ topology.

In general, the weak$^\ast$ and the weak topology do not coincide for measures on $\mathbb{R}^d$, but they do for measures on the compact $X$, as then $C_0^\ast(X)=C_b^\ast(X)$, where $C_b(X)$ is the space of all bounded and continuous functions on $X$. This means that the set $\mathcal{M}(X)$ of all probability measures on $X$ equipped with the total variation norm is weak$^*$- and therefore weakly compact.

On compact sets the $p$-Wasserstein distance (for $p\in[1,\infty)$) metrizes weak convergence (Theorem 5.10 in Santambrogio 2015), from which it follows that the set you defined is compact since it is closed.

In my opinion, a great reference for these sorts of questions is:

Santambrogio, F (2015): "Optimal Transport for Applied Mathematicians", Birkhäuser