When a charged particle accelerates through a potential difference, the potential energy acquired by the particle = work done in the process

                                              = charge(q) x potential difference(V).

                                              = qV

And this will be a increase in P.E. if the initial P.E. is lower than final one and in such a case there will be decrease in K.E.

But the condition in your problem states that the charges are accelerated. Thus the K.E. has increased and so P.E. must have decreased (i,e initial P.E. > final P.E.), and here we apply the principle of conservation of energy to state that--

     Loss in P.E. = Gain in K.E. 

  => qV = 1/2 mv² (v = gain in velocity starting from rest and m=mass of charge)

  => v =  (2qV/m)

i,e when V and m are same (as said in question) , v   q

Thus in your question

let velocity of +q charge be v₁ and that of +4q charge be v₂

thus v₁/v₂ =  (q/4q)

               =  (1/4)

               = 1/2

thus v₁ : v₂ = 1 : 2

thus option 4 is correct friend......