Kolbe Schmitt followed by methylation The first step in this synthesis is a Kolbe Schmitt process, but why is two equivalents of base needed? Also, what is the purpose of the two equivalents of methyl iodide? I assumed that there will be the protection of the phenol (to create a methyl ether), which would justify the two equivalents of base while converting the carboxylate into a methyl ester. However, the base is shown in separate steps, which makes me doubt if base would be available in the second step. Also, another problem is that if you follow the sequence (2 Grignard reagents used to convert the ester to a tertiary alcohol, followed by dehydration, followed by hydrogenation of the double bond just formed, followed by a Friedel Crafts acylation), you get to compound R, which only has 13 carbons, when if you had methylated the phenol, you would have 14. Does this mean that the hydrogenation somehow deprotected the methyl ether? Even more confusing, there is methylation occurring from R to S, which tells me that the phenol was protected at some point. How is this possible? 

EDIT: Here is the structure of the final product. The problem was about derivatives of natural products from Salvia plants, but I could not find a total synthesis of this compound anywhere. Also, there is no easy name for it so I couldn't search it well. If you can find the source, please let us know. Note that there are other steps in the synthesis that I didn't include in this question.
Collecting my comment thoughts together, there is an early demethylation step in this synthesis. Structure 1 is the result of carboxylation while methylation provides ester N. Friedel-Crafts acylations of anisoles at the ortho-position can undergo demethylation via aluminum chelation. Magnesium chelate 2 is a candidate for demethylation by iodide ion. Phenol O undergoes thermal dehydration catalyzed by the acidic phenolic hydroxyl group. Steps P --> Q are straightforward. Step R --> S methylates the phenol hydroxyl and excess KOH insures that any ester that is formed is saponified. Clemmensen reduction of S and remethylation afford T. 
Addendum: The OP's comment reveals that T is a carboxylic acid, not a methyl ester. Presumably the Clemmensen reduction effected some demethylation which is the reason for the last methylation step.