Prove NP-completeness of deciding satisfiability of monotone boolean formula
The "parent" of the problem you're looking at is sometimes called Weighted Satisfiability (WSAT, particularly in parameterized complexity) or Min-Ones (though this is normally an optimization version, but near enough). These problems have the "at most $k$ variables set to true" restriction as their defining feature.

The restriction to monotone formulae is actually surprisingly easy to show hardness for, you just need to thing outside satisfiability problems for a moment. Instead of trying to modify a SAT instance, we instead start with Dominating Set (DS).

See if you can get it from there. More is in the spoilers, broken into bits, but avoid them if you can. I won't show membership in NP, you should have no problem with that.


   Given an instance $(G, k)$ of DS (i.e. we want a dominating set of size at most $k$ for $G$), we can construct an instance $(\phi,k)$ of WSAT where the formula $\phi$ is a monotone CNF formula:


The basic construction:


   For each $v\in V(G)$ we have a variable $v' \in \text{var}(\phi)$, for each $v \in V(G)$ we have a clause $c_{v} = \bigvee_{u\in N(v)} u'$.


A sketch of the proof:


  Each vertex either has to be in the dominating set, or have a neighbour that is, so if we can find $k$ vertices that form a dominating set, the corresponding $k$ variables can be set to true in $\phi$, and each clause must contain at least one of them. Similarly if there is a weight $k$ satisfying assignment, the true variables correspond to the vertices we place in the dominating set - every clause $c_{v}$ must have at least one, so each $v$ is dominated (by itself or otherwise).