= Solution
A counterexample can even use a nonalgebraic <isolated type>. Let $L=\{<\}$, $\mathcal M=(\mathbb Q,<)$, $A=\varnothing$, and $P=\operatorname{tp}^{\mathcal M}(0/\varnothing)$. The <quantifier elimination for dense linear orders without endpoints> shows that all elements have the same <complete type> over the empty set: a one-variable <quantifier-free formula> has no order information except $x=x$ and $x<x$.
Thus $x=x$ isolates $P$. Every $\mathcal N\equiv\mathcal M$ is a <dense linear order without endpoints>, and every element of $N$ realizes $P$. Such a model cannot omit it. \b[Even a type with infinitely many realizations need not be omittable; isolation is the relevant obstruction.]
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