Given a 3SAT problem where each clause is a Monotone Clause (i.e. each clause with all positive or negative literals).
Moreover, we are promised that each solution for the above SAT problem (if any exists at all) will be of the form 1 in 3 SAT.
Monotone 1 in 3 SAT is NP Complete, but with the additional promise above, does the problem still remain NP Complete (it obviously is NP) ?
The trivial answer is no, as promise problems cannot be (by definition) in NP, and thus cannot be NP complete. The basic complexity classes e.g. $P, NP$ refer only to decision problems.
However, if you want to relate your promise problem to the standard classes, you can say it is NP-hard in the sense that if it can be solved in polynomial time for inputs in the promise, then $P=NP$.
To see why the above holds, let us introduce some notations. Let $C$ be the set of all monotone clause 3CNF, for which all satisfying assignments are 1 in 3. We shall call this set "the promise". Now your problem is, given an input in $C$, to determine whether it lies in $L_{yes}=C\cap SAT$, or in $L_{no}=C\setminus L_{yes}=C\cap\overline{SAT}$. Given an instance $\varphi$ to monotone clause 1 in 3 SAT, transform it to a monotone clause 3CNF $\psi$ such that $\psi\in C$, and in addition every 1 in 3 satisfying assignment for $\varphi$ also satisfies $\psi$. You can construct $\psi$ by adding clauses which negate conjunctions of pairs of literals in the same clause. Since $\psi\in C$ for every possible $\varphi$, and in addition $\varphi$ has a one in three satisfying assignment iff $\psi\in L_{yes}$, then solving the promise problem in polynomial time allows you to determine whether $\varphi$ has a one in three satisfying assignment, and thus $P=NP$.
The above basically shows that you can reduce an NP complete problem to your promise problem, but when talking about promise problems the definition of reduction should be slightly adjusted. Let us say that $f$ is a reduction from $L$ to the promise problem $L_{yes}\cup L_{no}$ if $Im(f)\subseteq L_{yes}\cup L_{no}$ and for every $x$ it holds that $x\in L\iff f(x)\in L_{yes}$. This is equivalent to saying that $x\in L\Rightarrow f(x)\in L_{yes}$ and $x\notin L\Rightarrow f(x)\in L_{no}$ (note that not being in $L_{yes}$ does not imply being in $L_{no}$).
