My attempt at this proof goes as follows: Consider the function $f$ paired with some partially ordered abstract operation like $\leq$. Since it is known to be a poset isomorphism, it must be bijective, and hence its inverse exists. Suppose for all elements $p$ in $P$, there exists some $p$ $\leq$ $f^-1$$(q)$. Then this bi-conditionally implies that for all $q$ in $Q$, there exists some $q=f(p)$ $\leq$ $q$. But this is the definition of a Galois Connection and hence we are done. Note that f: P->Q and its inverse: Q->P.
I don't think my proof is correct and need some assistance.
Suppose $p\le f^{-1}(q)$. Since $f$ is a poset homomorphism, this implies $$ f(p)\le f(f^{-1}(q))=q $$ Similarly for the other direction, because by definition of poset isomorphism, also $f^{-1}$ is a poset homomorphism.