In general, you can interpret let _ = <EXPR>; as execute <EXPR> and store the result, but I don't care about naming the storage place.
The net effect of the code you mention would be to turn the raw ptr back into an Rc so that there is no memleak, but at the same time the code doesn't care about using the resulting Rc itself. It only cares that the *const T -> Rc<T> conversion happens.
The main reason for the let _ = part would be to keep the compiler quiet. I don't think in this case the compiler would otherwise complain as the Rc type is not marked with the #[must_use] attribute, but if the <EXPR> had yielded a Result<X, Y> value (which is marked with #[must_use]), then the compiler would complain if nothing is done with the <EXPR> result value. Thus, the let _ = ... is a way to pre-empt that warning no matter the type.
Small nitpick, because I had the same misconception before: let _ = does not store anything, nor does it move the right-hand side. It's materially different from let _x =. Demo: Rust Playground