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.
Gödel/Scott themselves use both Ax. 2 and Ax. 3. The only difference is in Axiom 1 where you omit the box operator inside the scope of the allquantifier.
Here is why I think this is a troublesome assumption.
It might well be that it is accidentally the case that all individuals in our World that have a certain positive property A also have a Property B. For example: Suppose only humans have an appreciation for beauty, and that this is a positive property [our property A].
By Ax. 1 it would follow that every property B that all humans possess is also a positive property;
Simply because
"$ A(x) \rightarrow ishuman(x)$" and "$ishuman(x) \rightarrow B(x)$" implies
"$A(x) \rightarrow B(x)$"
This is a conclusion that is - to say the least - not very intuitive. That would mean that every contingent feature of human biology [from blood circulation to being able to be killed by a stick] is a positive, even godly property. Gödels formulation avoid these pitfalls and seems to be more plausible.
Best Answer
The modal operator $\square$ refers to necessity; its dual, $\lozenge$, refers to possibility. (A sentence is necessarily true iff it isn't possible for it to be false, and vice versa.) $P(\varphi)$ means that $\varphi$ is a positive (in the sense of "good") property; I'll just transcribe it as "$\varphi$ is good". I'll write out the argument colloquially, with the loss of precision that implies. In particular, the words "possible" and "necessary" are vague, and you need to understand modal logic somewhat to follow their precise usage in this argument.
Proof of Theorem $1$: Suppose $\varphi$ were good, but necessarily nothing had property $\varphi$. Then property $\varphi$ would, vacuously, force every other property; in particular $\varphi$ would force $\neg\varphi$. By Axiom $1$, this would mean that $\neg\varphi$ was also good; but this would then contradict Axiom $2$.
Proof of Theorem $2$: This follows directly from Theorem $1$ applied to Axiom $3$.
Proof of Theorem $3$: First note that if $x$ is godlike, it has all good properties (by definition) and no bad properties (by Axiom $2$). So any property that a godlike thing has is good, and is therefore necessarily good (by Axiom $4$), and is therefore necessarily possessed by anything godlike.
Proof of Theorem $4$: If something is godlike, it has every good property by definition. In particular, it's indispensable, since that's a good property (by Axiom $5$); so by definition something with its essence, which is just "being godlike" (by Theorem $3$), must exist. In other words, if something godlike exists, then it's necessary for something godlike to exist. But by Theorem $2$, it's possible that something godlike exists; so it's possible that it's necessary for something godlike to exist; and so it is, in fact, necessary for something godlike to exist. QED.
Convinced?