TLDR: When an object is fibrant in a model category, the endpoint maps for a very good path space object are fibrations. When is the converse true? That is, when can we conclude that an object is fibrant by knowing its endpoint maps are fibrations?
Details: Let $\mathcal M$ be a category with a model structure $(\mathcal C, \mathcal W, \mathcal F)$. For an object $X$, take a very good path space object to be a factorization of the diagonal $\Delta_X:X \rightarrow X \times X$ as $$X \xrightarrow{r} P_X \xrightarrow \epsilon X\times X$$ Where $r$ is an acyclic cofibration and $\epsilon$ is a fibration.
We think of $\epsilon$ as encoding the endpoints of a path in $X$. If $X$ is fibrant, you may easily verify that the projection maps $\pi_0, \pi_1: X \times X \rightarrow X$ are fibrations. It follows that $\pi_i \epsilon$ is a fibration for $i = 0,1$.
I’m curious about the converse to this. Suppose we know that $\pi_i \epsilon$ is a fibration for $i = 0,1$. Can we conclude that $X$ is fibrant? If not, are there model categories in which every object has a path space object with this property? Is this possible for, say, the standard model structure on simplicial sets?
I have not yet a good answer for arbitrary model categories, but as a partial answer I can prove for simplicial sets that if $\pi_1\varepsilon\colon P_X\to X$ is a Kan fibration, then $X$ is a Kan complex. I suspect that the same is true for Joyal fibrations and quasicategories.
So, we suppose given your setup in the Kan model structure on $\mathsf{sSet}$. We need to show that $X$ is a Kan complex, so suppose given a commutative diagram $$ \require{AMScd} \begin{CD} \Lambda^n_k @>{f}>>X\\ @VVV @VVV\\ \Delta^n@>>>* \end{CD} $$ for $n\geq 1$ and $0\leq k\leq n$. Write $f(0)=x$. Since $\varepsilon\colon P_X\to X\times X$ is a Kan fibration, so is its fiber $\varepsilon_x\colon P_{X,x}\to X\times\{x\}\cong X$, where we wrote $P_{X,x}=P_X\times_{X\times X}(X\times\{x\})$. Note that $x$ lies in the image of $\varepsilon_x$, because the diagonal of $X\times X$ lies in the image of $\varepsilon$. Say that $\widetilde{x}\in P_{X,x}$ satisfies $\varepsilon_x(\widetilde{x})=x$. Since $0\colon *\to\Lambda^n_k$ is a trivial cofibration, we can find in the commutative diagram $$ \require{AMScd} \begin{CD} *@>{\widetilde{x}}>> P_{X,x}\\ @V{0}VV @VV{\varepsilon_x}V\\ \Lambda^n_k @>{f}>>X\\ @VVV @VVV\\ \Delta^n@>>>* \end{CD} $$ a lift in the upper square. Since the $\pi_1\varepsilon\colon P_X\to X$ is assumed to be a Kan fibration, so is its pullback along $x\colon *\to X$. This means that the right vertical composite $P_{X,x}\to X\to *$ in the diagram above is a Kan fibration. Via standard lifting calculus, we obtain a lift in the lower square as well, so $X$ is a Kan complex.