(Community Wiki Answer)
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:
$$
\array{
X \times_{\Delta[1]} \{0\} &\to& X
\\
\downarrow && \downarrow
\\
\{0\} &\to& \Delta[1]
\\
\downarrow && \downarrow
\\
\{0\}^{\triangleright} &\to & P
}
$$
