= Solution
Let $\mathbf A$ be the one-element <partially ordered set> $\{a\}$ with $R(a,a)$. Taking $x=a$ witnesses the <existential quantifier>, and the only $y$ is comparable with $a$. Thus $\mathbf A\models\varphi$.
Let $\mathbf B$ be the two-element <antichain> $\{a,b\}$, with $R$ consisting only of $(a,a)$ and $(b,b)$. This is a <partial order>. If $x=a$, choose $y=b$, and if $x=b$, choose $y=a$. In either case both $R(x,y)$ and $R(y,x)$ are false. Therefore no witness for the <existential quantifier> exists:
$$
\boxed{\mathbf A=(\{a\},=)\models\varphi,\qquad\mathbf B=(\{a,b\},=)\not\models\varphi.}
$$
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