It is tempting to think this follows from a combination of Lemmas \ref{lemma-bottom-part-degenerates} and \ref{lemma-duality-hodge}. However this doesn't work because the maps $\mathcal{O}_X \to \Omega^1_{X/k}$ and $\Omega^{d - 1}_{X/k} \to \Omega^d_{X/k}$ are not $\mathcal{O}_X$-linear and hence we cannot use the functoriality discussed in Duality for Schemes, Remark \ref{duality-remark-coherent-duality-proper-over-field} to conclude the map in Lemma \ref{lemma-bottom-part-degenerates} is dual to the one in this lemma. \medskip\noindent We may replace $X$ by a connected component of $X$. Hence we may assume $X$ is irreducible. By Varieties, Lemmas \ref{varieties-lemma-smooth-geometrically-normal} and \ref{varieties-lemma-proper-geometrically-reduced-global-sections} we see that $k' = H^0(X, \mathcal{O}_X)$ $X$ is a finite separable extension $k'/k$. Since $\Omega_{k'/k} = 0$ (see for example Algebra, Lemma \ref{algebra-lemma-characterize-separable-algebraic-field-extensions}) we see that $\Omega_{X/k} = \Omega_{X/k'}$ (see Morphisms, Lemma \ref{morphisms-lemma-triangle-differentials}). Thus we may replace $k$ by $k'$ and assume that $H^0(X, \mathcal{O}_X) = k$. \medskip\noindent Assume $H^0(X, \mathcal{O}_X) = k$. We conclude that $\dim H^d(X, \Omega^d_{X/k}) = 1$ by Lemma \ref{lemma-duality-hodge}. Assume first that the characteristic of $k$ is a prime number $p$. Denote $F_{X/k} : X \to X^{(p)}$ the relative Frobenius of $X$ over $k$; please keep in mind the facts proved about this morphism in Lemma \ref{lemma-Garel-map-frobenius-smooth-char-p}. Consider the commutative diagram $$ \xymatrix{ H^d(X, \Omega^{d - 1}_{X/k}) \ar[d] \ar[r] & H^d(X^{(p)}, F_{X/k, *}\Omega^{d - 1}_{X/k}) \ar[d] \ar[r]_{\Theta^{d - 1}} & H^d(X^{(p)}, \Omega^{d - 1}_{X^{(p)}/k}) \ar[d] \\ H^d(X, \Omega^d_{X/k}) \ar[r] & H^d(X^{(p)}, F_{X/k, *}\Omega^d_{X/k}) \ar[r]^{\Theta^d} & H^d(X^{(p)}, \Omega^d_{X^{(p)}/k}) } $$ The left two horizontal arrows are isomorphisms as $F_{X/k}$ is finite, see Cohomology of Schemes, Lemma \ref{coherent-lemma-relative-affine-cohomology}. The right square commutes as $\Theta_{X^{(p)}/X}$ is a morphism of complexes and $\Theta^{d - 1}$ is zero. Thus it suffices to show that $\Theta^d$ is nonzero (because the dimension of the source of the map $\Theta^d$ is $1$ by the discussion above). However, we know that $$ \Theta^d : F_{X/k, *}\Omega^d_{X/k} \to \Omega^d_{X^{(p)}/k} $$ is surjective and hence surjective after applying the right exact functor $H^d(X^{(p)}, -)$ (right exactness by the vanishing of cohomology beyond $d$ as follows from Cohomology, Proposition \ref{cohomology-proposition-vanishing-Noetherian}). Finally, $H^d(X^{(d)}, \Omega^d_{X^{(d)}/k})$ is nonzero for example because it is dual to $H^0(X^{(d)}, \mathcal{O}_{X^{(p)}})$ by Lemma \ref{lemma-duality-hodge} applied to $X^{(p)}$ over $k$. This finishes the proof in this case. \medskip\noindent Finally, assume the characteristic of $k$ is $0$. We can write $k$ as the filtered colimit of its finite type $\mathbf{Z}$-subalgebras $R$. For one of these we can find a cartesian diagram of schemes $$ \xymatrix{ X \ar[d] \ar[r] & Y \ar[d] \\ \Spec(k) \ar[r] & \Spec(R) } $$ such that $Y \to \Spec(R)$ is smooth of relative dimension $d$ and proper. See Limits, Lemmas \ref{limits-lemma-descend-finite-presentation}, \ref{limits-lemma-descend-smooth}, \ref{limits-lemma-descend-dimension-d}, and \ref{limits-lemma-eventually-proper}. The modules $M^{i, j} = H^j(Y, \Omega^i_{Y/R})$ are finite $R$-modules, see Cohomology of Schemes, Lemma \ref{coherent-lemma-proper-over-affine-cohomology-finite}. Thus after replacing $R$ by a localization we may assume all of these modules are finite free. We have $M^{i, j} \otimes_R k = H^j(X, \Omega^i_{X/k})$ by flat base change (Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}). Thus it suffices to show that $M^{d - 1, d} \to M^{d, d}$ is zero. This is a map of finite free modules over a domain, hence it suffices to find a dense set of primes $\mathfrak p \subset R$ such that after tensoring with $\kappa(\mathfrak p)$ we get zero. Since $R$ is of finite type over $\mathbf{Z}$, we can take the collection of primes $\mathfrak p$ whose residue field has positive characteristic (details omitted). Observe that $$ M^{d - 1, d} \otimes_R \kappa(\mathfrak p) = H^d(Y_{\kappa(\mathfrak p)}, \Omega^{d - 1}_{Y_{\kappa(\mathfrak p)}/\kappa(\mathfrak p)}) $$ for example by Limits, Lemma \ref{limits-lemma-higher-direct-images-zero-above-dimension-fibre}. Similarly for $M^{d, d}$. Thus we see that $M^{d - 1, d} \otimes_R \kappa(\mathfrak p) \to M^{d, d} \otimes_R \kappa(\mathfrak p)$ is zero by the case of positive characteristic handled above.