Is it possible to prove $p\rightarrow\diamond (p\land q)$ in modal logic?

Question

I need to prove $p\rightarrow\diamond (p\land q)$ in B axiomatic, which contains next conversion rules:
1.$(p\land q)\rightarrow(q\land p)$
2.$(q\land p)\rightarrow p$
3.$p\rightarrow(p\land p)$
4.$p\land(q\land r)\rightarrow(p\land q)\land r$
5.$p\rightarrow \lnot(\lnot p)$
6.$(p\land q)\land(q\land r)\rightarrow(p\land r)$
7.$(p\land(p\rightarrow q))\rightarrow q$
8.$\diamond(p\land q)\rightarrow\diamond p$

Answer 1

This answers the question as it was before it was edited to ask something completely different!

If those are all the axioms you have, then no. The axioms are all true in an interpretation where $\Diamond p$ is false for all $p$, and in such an interpretation $p\to\Diamond(p\land q)$ will be false whenever $p$ is true.

On the other hand, it looks rather unlikely that those are really all the axioms you have. Most of them look like a randomish collection of algebraic properties of conjunction, but there seems to be nothing that allows you to reason much about implication. And apparently the only thing you can do with negation there is to introduce double negations; none of your rules say anything about negations that aren't doubled.

Are you really sure that's all of your rules?