Is a Methoxy group meta directing in this problem?
Now that the general consensus of the Commenters is that meta cyclization is viable and that the answer is (d) and not (a), the OP and other readers should be concerned as to how an (E)-double bond can cyclize in a Friedel-Crafts (F-C) reaction. It doesn't! Prior isomerization to the (Z)-configuration is required. 
(E)-Acyl chloride 1 must rotate about the red β-γ single bond and $\ce{AlCl3}$ must complex with the carbonyl oxygen forming species 2. [One may also invoke $\ce{AlCl3}$ complexation with the chlorine and rationalize the subsequent steps via an acylium ion and conjugated ketenes. The principle is the same.] Loss of a proton at the γ-position of conformation 2 provides the β,γ-(Z)-double bond of 3. Now deconjugated acyl chloride 4 is ready for F-C cyclization to intermediate 7 and onto 7-methoxynaphthalen-1-ol (9), aka (d). Alternatively, rotation about the red bond in 3 leads to conformation 5, which, upon γ-protonation leads to α,β-(Z)-conjugated acyl chloride 6. Subsequent F-C cyclization via intermediate 8 affords 9. Incidently, base is not required to produce 9 from 7 or 8. F-C conditions are sufficient.