= Solution
Write $L\alpha=\omega\wedge\alpha$, the <Lefschetz operator of a Kähler manifold>. The metric and volume form define the $L^2$ <inner product> on smooth complex forms, and $\Lambda=L^*$ is its <formal adjoint>. Similarly, $\bar\partial^*$ is the <formal adjoint> of the <Dolbeault operator>, characterized by $\langle\bar\partial\beta,\alpha\rangle=\langle\beta,\bar\partial^*\alpha\rangle$. The <Dolbeault Laplacian> is
$$
\boxed{\Delta_{\bar\partial}=\bar\partial\bar\partial^*+\bar\partial^*\bar\partial.}
$$
On the compact manifold without boundary, integration by parts gives
$$
\langle\Delta_{\bar\partial}\alpha,\alpha\rangle
=\|\bar\partial\alpha\|_2^2+\|\bar\partial^*\alpha\|_2^2.
$$
If the Laplacian vanishes, both terms are zero. Conversely, if both operators annihilate $\alpha$, the defining formula annihilates it. Thus \b[harmonicity is equivalent to being both $\bar\partial$-closed and $\bar\partial^*$-closed].
Because $\omega$ is closed and has type $(1,1)$, $\bar\partial\omega=0$ and $[L,\bar\partial]=0$. The supplied identity from the <Kähler identities> gives $[L,\bar\partial^*]=-i\partial$. With ordinary commutators for the even-degree operator $L$,
$$
\begin{aligned}
[L,\Delta_{\bar\partial}]&=[L,\bar\partial]\bar\partial^*
+\bar\partial[L,\bar\partial^*]+[L,\bar\partial^*]\bar\partial
+\bar\partial^*[L,\bar\partial]\\
&=-i(\bar\partial\partial+\partial\bar\partial)=0.
\end{aligned}
$$
The last identity follows from $d^2=0$. This also proves that the Lefschetz operator preserves harmonic forms.
For the cohomology map one can work directly with forms: $\bar\partial(\omega^k\wedge\alpha)=\omega^k\wedge\bar\partial\alpha$, since $\omega$ has even degree. It takes closed forms to closed forms and exact forms to exact forms. Therefore the $k$th power of $L$ induces
$$
\boxed{\phi_{\omega,k}([\alpha])=[\omega^k\wedge\alpha].}
$$
The bidegree is $(p+k,q+k)$, including the zero groups outside the dimension range. No isomorphism claim is needed here.
Finally put $\theta=-i\partial f$, so $\bar\partial\theta=i\partial\bar\partial f=\omega'-\omega$. Let
$$
T_k=\sum_{j=0}^{k-1}(\omega')^j\wedge\omega^{k-1-j}.
$$
Both <Kähler forms> are $\bar\partial$-closed, so $\bar\partial T_k=0$. For a closed representative $\alpha$,
$$
(\omega')^k\wedge\alpha-\omega^k\wedge\alpha
=\bar\partial\bigl(\theta\wedge T_k\wedge\alpha\bigr).
$$
This primitive has type $(p+k,q+k-1)$; the degree-one sign produces no additional term because $T_k\wedge\alpha$ is closed. Hence
$$
\boxed{\phi_{\omega,k}=\phi_{\omega',k}\quad\text{for every }k\geq1.}
$$
This is the <dependence of Lefschetz maps on the Dolbeault class>.
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