It has to do with line notation of an EC cell:

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so to answer your question we just apply what we already know from Le Chatlier --> if I increase Reactants I will make more product.

Reactants = solid lead & aqueous copper ... so if increase this then i will get more of my intended product which is current

Their option C is wrong then. I hope someone somewhere was helped by this !! Berkeley Review helped me a ton for C/P and i initially struggled with this concept a ton. GLWS

Actually, this question has everything to do with the Nernst equation. In addition, there is a major flaw in your argument. Pure solids and liquids do not experience concentration changes and thus have no effect on Le Chateliers principle/ on the equilibrium position of the reaction. In other words, you could add more solid, and it would have no effect. In addition, they only gave a generalized X+ and Y+. we have no clue as to what is happening at the anode or cathode, thus, we have no clue what our reactants are and what our products are. All we know is that, solids are not included in the reaction quotient, thus, Q=[X+]/[Y+] or Q= [Y+]/[X+] depending on what is oxidized and what is reduced. We can then use the Nernst equation to see how this would affect the voltage and thus the current.

Nernst equation is used for electrochemical cells that are not under standard conditions (ie. we changed the concentration). The Nernst equation is added in a picture below. If i decrease both ionic species (I assumed that they decreased them by the same amount), my reaction quotient should remain the same and thus my cell potential (Ecell) should also remain the same. So to answer OP's question, I think that the E cell and thus voltage would remain the same and thus there would be no change in the current through the system. Let me know if you agree!

Hope that helps! good luck