Urs Schreiber answered this question over at the nForum, so I'll reproduce his answer here (since I don't think he's going to add it himself)

Since $\mathfrak{C}$ is left adjoint we can essentially compute the pushout before applying $\mathfrak{C}$. Let me call the analog of $M$ obtained this way $P$

$$\array{ X &\to& X^{\triangleright} \\ \downarrow && \downarrow \\ \Delta[1] &\to& P }$$

We have a canonical map $P \to \Delta[1]^{\triangleright}$ induced from the commutativity of

$$\array{ X &\to& X^{\triangleright} \\ \downarrow && \downarrow \\ \Delta[1] &\to& \Delta[1]^{\triangleright} } \,.$$

For evaluating $P(0,p)$ we just need the fiber over ${0}^{\triangleright}$, hence the pullback of the diagram

$$\array{ && P \\ && \downarrow \\ \{0\}^{\triangleright} &\hookrightarrow& \Delta[1]^{\triangleright} } \,.$$

Now, since colimits commute with pullbacks in $sSet$, this pullback is the pushout of the corresponding pullbacks of $X$, and $X^{\triangleright}$. But that pullback of $X$ is $X \times_{\Delta[1]} \Delta[0]$. Because you can compute it as this consecutive pullback:

$$\array{ X \times_{\Delta[1]} \{0\} &\to& X \\ \downarrow && \downarrow \\ \{0\} &\to& \Delta[1] \\ \downarrow && \downarrow \\ \{0\}^{\triangleright} &\to & P }$$