Yes. If you are given a simplicial set $X: \Delta^{\text{op}} \to \text{Sets}$, then the the thick realization $||X||$ of $X$ is given by the same formula as the ordinary realization with the exception that one only uses injective order preserving maps of finite ordered sets. The map $||X|| \to |X|$ is always a homotopy equivalence (here's a reference: https://ncatlab.org/nlab/show/geometric+realization+of+simplicial+topological+spaces#GoodAndProper).

Now, given an ordered simplicial complex $K$, the promotion $Ss(K)$ of $K$ to a simplicial set has the property that $||Ss(K)|| = |K|$ identically, where $|K|$ means the geometric realization of $K$ as a simplicial complex. So, $$ |K| = ||Ss(K)|| \simeq |Ss(K)|\, . $$.

Edit: the second paragraph isn't correct as stated. It isn't true that $||Ss(K)|| = |K|$ identically as Neil points out. I am going to keep this post here until I can find a correction.

Yes. If you are given a simplicial set $X: \Delta^{\text{op}} \to \text{Sets}$, then the the thick realization $||X||$ of $X$ is given by the same formula as the ordinary realization with the exception that one only uses injective order preserving maps of finite ordered sets. The map $||X|| \to |X|$ is always a homotopy equivalence (here's a reference: https://ncatlab.org/nlab/show/geometric+realization+of+simplicial+topological+spaces#GoodAndProper).

Now, given an ordered simplicial complex $K$, the promotion $Ss(K)$ of $K$ to a simplicial set has the property that $||Ss(K)|| = |K|$ identically, where $|K|$ means the geometric realization of $K$ as a simplicial complex. So, $$ |K| = ||Ss(K)|| \simeq |Ss(K)|\, . $$.

Edit: the second paragraph isn't correct as stated. It isn't true that $||Ss(K)|| = |K|$ identically as Neil points out. I am going to keep this post here until I can find a correction.

Correction: If $X$ is a simplicial set, we can consider the poset $P_X$ of its non-degenerate simplices. The nerve of this poset is the barycentric subdivision of $X$. If $X = Ss(K)$, where $K$ is as above, then $P_X$ coincides with the poset of simplices of $K$ under inclusion.

Therefore, with $X= Ss(K)$, the nerve of the poset $P_X$ coincides with the barycentric subdivision of $K$, so by homotopy invariance of the subdivision, we see that $|X| \simeq |P_X| \simeq |K|$.

Yes. If you are given a simplicial set $X: \Delta^{\text{op}} \to \text{Sets}$, then the the thick realization $||X||$ of $X$ is given by the same formula as the ordinary realization with the exception that one only uses injective order preserving maps of finite ordered sets. The map $||X|| \to |X|$ is always a homotopy equivalence (here's a reference: https://ncatlab.org/nlab/show/geometric+realization+of+simplicial+topological+spaces#GoodAndProper).

Now, given an ordered simplicial complex $K$, the promotion $Ss(K)$ of $K$ to a simplicial set has the property that $||Ss(K)|| = |K|$ identically, where $|K|$ means the geometric realization of $K$ as a simplicial complex. So, $$ |K| = ||Ss(K)|| \simeq |Ss(K)|\, . $$.

Yes. If you are given a simplicial set $X: \Delta^{\text{op}} \to \text{Sets}$, then the the thick realization $||X||$ of $X$ is given by the same formula as the ordinary realization with the exception that one only uses injective order preserving maps of finite ordered sets. The map $||X|| \to |X|$ is always a homotopy equivalence (here's a reference: https://ncatlab.org/nlab/show/geometric+realization+of+simplicial+topological+spaces#GoodAndProper).

Now, given an ordered simplicial complex $K$, the promotion $Ss(K)$ of $K$ to a simplicial set has the property that $||Ss(K)|| = |K|$ identically, where $|K|$ means the geometric realization of $K$ as a simplicial complex. So, $$ |K| = ||Ss(K)|| \simeq |Ss(K)|\, . $$.

Edit: the second paragraph isn't correct as stated. It isn't true that $||Ss(K)|| = |K|$ identically as Neil points out. I am going to keep this post here until I can find a correction.