= Solution
Write $L=W_p(\mu,\nu)$. The assumed equality of the <Monge optimal transport problem> and <Kantorovich optimal transport problem> values gives
$$
\int|T^\dagger(x)-x|^p\,d\mu(x)=L^p.
$$
Each interpolated <pushforward measure> $\mu_t=(P_t)_\#\mu$ has a finite $p$th <absolute moment>, since $|P_t(x)|^p\leq2^{p-1}(|x|^p+|T^\dagger(x)|^p)$ and $(T^\dagger)_\#\mu=\nu$.
For any $s,t\in[0,1]$, the common-source <transport plan>
$$
\pi_{s,t}=(P_s,P_t)_\#\mu\in\Pi(\mu_s,\mu_t)
$$
provides the upper bound
$$
W_p(\mu_s,\mu_t)^p\leq\int|P_t(x)-P_s(x)|^p\,d\mu
=|t-s|^p L^p.
$$
For the reverse bound assume $0\leq s\leq t\leq1$. Since $\mu_0=\mu$ and $\mu_1=\nu$, the <triangle inequality> for the <p-Wasserstein distance> and the upper bounds already established give
$$
\begin{aligned}
L&\leq W_p(\mu_0,\mu_s)+W_p(\mu_s,\mu_t)+W_p(\mu_t,\mu_1)\\
&\leq sL+W_p(\mu_s,\mu_t)+(1-t)L.
\end{aligned}
$$
Hence $W_p(\mu_s,\mu_t)\geq(t-s)L$. Combining the bounds and using symmetry proves
$$
\boxed{W_p(\mu_t,\mu_s)=|t-s|W_p(\mu,\nu).}
$$
This is the constant speed property of <displacement interpolation>. The given invertibility of $P_s$ also lets one realize the competitor as the map $P_t\circ P_s^{-1}$, assuming its inverse is measurable, but the <transport plan> argument proves the result without invertibility.
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