Lemma: Let $\Delta := \{\delta_{x} \mid x \in X\}$. Then $\Delta$ is closed in $\mathcal P$.
Proof: Let $(\delta_{x_n})$ be a sequence in $\Delta$ with $x_1, x_2, \ldots \in X$ such that $\delta_{x_n} \to \mu \in \mathcal M$ in $d_P$. I proved here that $\mathcal P$ is closed in $\mathcal M$. So $\mu \in \mathcal P$.
WLOG, we assume that $d_P(\delta_{x_n}, \delta_{x_m}) <1/2$ for all $m,n$. By this result, we get $d(x_m, x_n) = d_P(\delta_{x_m}, \delta_{x_n})$. So $(x_n)$ is a Cauchy sequence.
Let $C_m := \overline{\{x_n \mid n \ge m\}}$ and $C :=\bigcap_m C_m$. Convergence in $d_P$ implies weak convergence, so
$$
1=\limsup_n \delta_{x_n} (C_m) \le \mu(C_m) \quad \forall m.
$$
It follows that $\mu(C_m)=1$ for all $m$, so $\mu(C)=1$. As such, $C \neq \emptyset$. By construction, $C$ is a singleton, i.e., $\exists x^* \in X$ such that $C=\{x^*\}$. Then $\mu = \delta_{x^*} \in \Delta$. This completes the proof.
Assume $\mathcal P$ is complete. Notice that $x \mapsto \delta_{x}$ is a homeomorphism from $X$ onto $\left\{\delta_{x} \mid x \in X\right\}$. Assume that $(x_n)$ is a Cauchy sequence in $X$. WLOG, we assume that $d(x_n, x_m) \le 1/2$ for all $n,m$. On the other hand, $d_P(\delta_{x_n}, \delta_{x_m}) = \min \{d(x_n, x_m), 1\} = d(x_n, x_m)$. It follows that $(\delta_{x_n})$ is a Cauchy sequence, so there is $\mu \in \mathcal P$ such that $\delta_{x_n} \to \mu \in \mathcal P$ in $d_P$. By our Lemma, $\Delta$ is closed in $\mathcal P$, so there is $x^*\in X$ such that $\mu = \delta_{x^*}$. Clearly, $x_n \to x^*$.
Assume $\mathcal M$ is complete. By this result, $\mathcal P$ is closed in $\mathcal M$, so $\mathcal P$ is complete. As we proved above, $X$ is complete.
