Let $K$ be a simplicial complex with vertex set $V$. Let $S_\bullet (K)$ be the simplicial set whose p-simplices are the maps $f:[p]\to V$ such that $f([p])$ is a simplex of $K$, or alternatively the set of maps $\Delta^p \to K$ of simplicial complexes. There is an obvious map $$\pi_K:|S_\bullet (K)| \to |K|$$ which you ask to be a homotopy equivalence. Here is an argument. I find it easier to work with the fat geometric realization $||S_\bullet (K)||$ instead, but the difference is minimal, since the quotient map to the ordinary geometric realization is a homotopy equivalence.

Step 1. Consider first the case $K=\Delta^n$ (rather, it is the full simplicial complex with vertex set $[n]$). I claim that $||S_\bullet \Delta^n||$ is contractible. For sake of notational clarity, let me write $\nabla^p$ for the topological $p$-simplex. Consider the map $$H_p: S_p (\Delta^n) \times \nabla^p \times [0,1] \to S_{p+1}(\Delta^n) \times \nabla^{p+1} $$ which is given by the formula $$H(f,v,t):= (f \ast n,((1-t)v,t)). $$ Explanation: $f \ast n: [p+1] \to [n]$ is the map whose restriction to $[p]$ is $f$ and which has $f(p+1)=n$. Furthermore $((1-t)v,t) \in \mathbb{R}^{p+1} \times \mathbb{R}$ is a point of $\nabla^{p+1}$. It is easily checked that the different $H_p$ glue together to a map $H:|| S_\bullet (\Delta^n)|| \times [0,1] \to ||S_{\bullet}(\Delta^n) ||$ (use that products and quotients commute in this setting, as the interval is compact, or work in the context of compactly generated spaces). It is clear that $H(0,\_)$ is the identity, and $H(1,\_)$ is the constant map to the vertex $n$. So we are done in this case.

Step 2. Now we prove the claim for finite complexes, by induction over both, the dimension and the number of top-dimensional simplices. The induction beginning $K=\emptyset$ is trivial. For the induction step, let $K$ be $n$-dimensional and let $L$ be obtained from $K$ by deleting one $n$-simplex. Then $|K| \cong |L| \cup_{|\partial \Delta^n|} |\Delta^n|$ and $||S_\bullet (K)|| \cong ||S_\bullet (L)|| \cup_{||S_\bullet (\partial \Delta^n)||} ||S_\bullet (\Delta^n)||$. The map $\pi_K$ is the pushout of the maps $\pi_{\Delta^n}$ and $\pi_L$, along $\pi_{\partial \Delta^n}$. These maps are homotopy equivalences, by step 1 and by induction hypothesis, respectively. The maps $|\partial \Delta^n| \to |\Delta^n|$ and $||S_\bullet (\partial \Delta^n)|| \to ||S_\bullet (\Delta^n)||$ are cofibrations, and so the gluing lemma implies that $\pi_K$ is a homotopy equivalence.

Step 3. Having shown the claim for finite complexes, it follows by a colimit argument that $\pi_K$ is a weak homotopy equivalence for arbitrary $K$, and hence a homotopy equivalence, by Whiteheads theorem.

$\newcommand\geom[1]{\lvert#1\rvert}\newcommand\Geom[1]{\lVert#1\rVert}$Let $K$ be a simplicial complex with vertex set $V$. Let $S_\bullet (K)$ be the simplicial set whose $p$-simplices are the maps $f:[p]\to V$ such that $f([p])$ is a simplex of $K$, or alternatively the set of maps $\Delta^p \to K$ of simplicial complexes. There is an obvious map $$\pi_K:\geom{S_\bullet (K)} \to \geom K$$ which you ask to be a homotopy equivalence. Here is an argument. I find it easier to work with the fat geometric realization $\Geom{S_\bullet (K)}$ instead, but the difference is minimal, since the quotient map to the ordinary geometric realization is a homotopy equivalence.

Step 1. Consider first the case $K=\Delta^n$ (rather, it is the full simplicial complex with vertex set $[n]$). I claim that $\Geom{S_\bullet \Delta^n}$ is contractible. For sake of notational clarity, let me write $\nabla^p$ for the topological $p$-simplex. Consider the map $$H_p: S_p (\Delta^n) \times \nabla^p \times [0,1] \to S_{p+1}(\Delta^n) \times \nabla^{p+1} $$ which is given by the formula $$H(f,v,t):= (f \ast n,((1-t)v,t)). $$ Explanation: $f \ast n: [p+1] \to [n]$ is the map whose restriction to $[p]$ is $f$ and which has $f(p+1)=n$. Furthermore $((1-t)v,t) \in \mathbb{R}^{p+1} \times \mathbb{R}$ is a point of $\nabla^{p+1}$. It is easily checked that the different $H_p$ glue together to a map $H:\Geom{S_\bullet (\Delta^n)} \times [0,1] \to \Geom{S_{\bullet}(\Delta^n)}$ (use that products and quotients commute in this setting, as the interval is compact, or work in the context of compactly generated spaces). It is clear that $H(0,\_)$ is the identity, and $H(1,\_)$ is the constant map to the vertex $n$. So we are done in this case.

Step 2. Now we prove the claim for finite complexes, by induction over both, the dimension and the number of top-dimensional simplices. The induction beginning $K=\emptyset$ is trivial. For the induction step, let $K$ be $n$-dimensional and let $L$ be obtained from $K$ by deleting one $n$-simplex. Then $\geom K \cong \geom L \cup_{\geom{\partial \Delta^n}} \geom{\Delta^n}$ and $\Geom{S_\bullet (K)} \cong \Geom{S_\bullet (L)} \cup_{\Geom{S_\bullet (\partial \Delta^n)}} \Geom{S_\bullet (\Delta^n)}$. The map $\pi_K$ is the pushout of the maps $\pi_{\Delta^n}$ and $\pi_L$, along $\pi_{\partial \Delta^n}$. These maps are homotopy equivalences, by step 1 and by induction hypothesis, respectively. The maps $\geom{\partial \Delta^n} \to \geom{\Delta^n}$ and $\Geom{S_\bullet (\partial \Delta^n)} \to \Geom{S_\bullet (\Delta^n)}$ are cofibrations, and so the gluing lemma implies that $\pi_K$ is a homotopy equivalence.

Step 3. Having shown the claim for finite complexes, it follows by a colimit argument that $\pi_K$ is a weak homotopy equivalence for arbitrary $K$, and hence a homotopy equivalence, by Whitehead's theorem.