Check that in your frame you can describe in first order language that there are three types elements: Those that do not contain any elements (first type), those that contain elements that do not contain elements (second type) and a unique element that contains all the elements of the second type (third type).
Next check that via first order language you can make sure that the elements of the first type are infinite many, that all the elements of the second type contain infinite many elements and are closed under finite unions and intersections.
Now take a frame $(W,R)$ (I will assume w.l.o.g. that $R$ is ''$\ni$'' to improve the notation a bit) elementarily equivalent to your original frame and assume that it is countable. Let $A$ be the set of the countable elements of the first type and let $\{w_1,\ldots,w_n,\ldots\}$ be the set of the elements of the second type. Then I will construct a valuation such that the element of the third type falsifies the formula.
To do this pick $a_1,b_1\in w_1\cap A$ and let $a_1\in V(q)$ and $b_1\notin V(q)$. Let's define $A_1=A\setminus\{a_1,b_1\}$. Continue inductively: Assume you have a cofinite set $A_k$ and you have defined the valuation for the elements of $A\setminus A_k$ such that for every $m\leq k$ there are some $a,b\in A\setminus A_k$ such that $a,b\in w_m$ and $a\in V(q)$ while $b\notin V(q)$. Then since $w_{k+1}$ is infinite (by elementary equivalence) we have that $w_{k+1}\cap A_k$ is infinite. Hence we can pick $a_{k+1},b_{k+1}\in w_{k+1}\cap A_k$ and let $a_{k+1}\in V(q)$ while $b_{k+1}\notin V(q)$ and let $A_{k+1}=A_k\setminus\{a_{k+1},b_{k+1}\}$. Also let $V(p)=W$ and for define $V(q)$ arbitrarily elsewhere (after the inductive construction).
If $w$ is the unique element of the third type in your model you have that $(W,R,V),w\Vdash\diamondsuit\square p$ but $(W,R,V),w\nVdash\diamondsuit(\square(p\land \lnot q)\lor\square(p\land q))$. This is because for every successor of $w$, there is a successor of it that satisfies $q$ and one that doesn't.
I think that the construction would break down at the second step.
If a given logic is not compact, then there is a set $\Gamma$ every finite subset of which is satisfiable, but $\Gamma$ itself is not. Therefore, while proving Lindenbaum Lemma, we cannot be sure that the union of countably many consistent sets $\Gamma = \bigcup_{n=0}^\infty \Gamma_n$ is satisfiable.
Classic approaches to tackle this problem include using a finite fragment of a language via some kind of a closure (PDL, KL), or putting additional requirements on maximal consistent sets (e.g. that there is always a witness for a negation of a quantified formula of Quantified ML).
Best Answer
In the online copy of the book you linked to, the statement is $\square p\Vdash^g_{\mathsf{M(K)}} p$. If you have a print copy of the book that reads $p\Vdash^g_{\mathsf{M(K)}} \lozenge p$, maybe this is a typo that was corrected in the online version?