$(\Box p \wedge \Box q) \rightarrow \Box(p\wedge q)$ is valid in K. But $(\Diamond p \wedge \Diamond q) \rightarrow \Diamond(p\wedge q)$ is not.
I'm not sure what a frame that validates $(\Diamond p \wedge \Diamond q) \rightarrow \Diamond(p\wedge q)$ would look like. I wonder if $\Diamond$ reduces to $\Box$ in such a system? I'd appreciate any help on this.
Intuitively, $$\mbox{"$(\Diamond p\wedge\Diamond q)\rightarrow\Diamond(p\wedge q)$ is valid"}$$ says that I can never partition the worlds I see into two different (nonempty) pieces: if I could, consider making $p$ true and $q$ false on one half of the partition and $p$ false and $q$ true on the other half of the partition.
So suppose $\mathfrak{F}$ is a frame validating $(\Diamond p\wedge\Diamond q)\rightarrow\Diamond(p\wedge q)$; how many worlds can each world in $\mathfrak{F}$ possibly see?
And in fact it's easy to show that this is an exact characterization.