For which k is PLANAR NAE k-SAT in P? The Not All Equal $k$-SAT problem (NAE $k$-SAT), given a set $C$ of clauses over a set $X$ of boolean variables such that each clause contains at most $k$ literals, asks whether there exists a truth assignment of the variables such that each clause contains at least one true and at least one false literal.

The PLANAR NAE $k$-SAT problem is the restriction of NAE $k$-SAT to those instances where the incidence bipartite graph of $C$ and $X$ (i.e. the graph of parts $C$ and $X$ with an edge between $x\in X$ and $c\in C$ if and only if $x$ or $\overline{x}$ belongs to $c$), is planar.

It is known that NAE 3-SAT is NP-complete (Garey and Johnson, Computers and Intractability; A Guide to the Theory of NP-Completeness), but PLANAR NAE 3-SAT is in P (see Planar NAE3SAT is in P, B. Moret, ACM SIGACT News, Volume 19 Issue 2, Summer 1988 - unfortunately I do not have access to this paper).

Is PLANAR NAE $k$-SAT in P for some $k\geq 4$? Is there a value of $k$ for which it has been shown to be NP-complete?
PLANAR NAE $k$-SAT is in P for all values of $k$ . 

Reason is that we can reduce PLANAR NAE $k$-SAT to PLANAR NAE $3$-SAT. Let $\phi$ be an instance of PLANAR NAE $k$-SAT, and suppose $\phi$ contains a clause $C$ with literals $\ell_1, \ell_2, \dots, \ell_k$. Introduce a new variable $v_C$, and replace $C$ with two NAE clauses $C_1$ and $C_2$. $C_1$ contains $3$ literals $\ell_1$, $\ell_2$, and $v_C$, while $C_2$ contains $k-1$ literals $\bar{v}_C, \ell_3, \ell_4, \dots, \ell_k$. It's easy to see that $C$ is satisfiable iff $C_1 \wedge C_2$ is and that the transformation preserves planarity. Now, we can repeatedly apply this procedure on the clauses to eventually obtain an instance $\phi&#39;$ of NAE $3$-SAT as desired.