$\large I^{st}\space part$
$M[1,2] = M[1,1] + M[2,2] + P_0P_1P_2 = 2\times 25\times 3= 150 $
$M[2,3] = P_1P_2P_3 = 25\times 3\times16=1200$
$M[3,4] = P_2P_3P_4 = 3\times 16\times1=48$
$M[4,5] = P_3P_4P_5 = 16\times 1\times1000=16000$
$\large II^{nd}\space part$
In $M[1,3] $ $M[1,2]$ is minimum so
$M[1,3] = M[1,2] + M[3,3] + P_0P_2P_3 = 150 + 2\times 3\times 16=243$
so minimum is $M[3,4]$ so
$M[2,4] = M[2,2] + M[3,4] + P_1P_2P_4 = 48 + 25\times3\times1 = 123 $
now compute further and $M[2,4] $ is minimum so
in $M[3,5]$ $M[3,4]$ is already minimum so
$M[3,5] = M[3,4] + M[5,5] + P_2P_4P_5 = 48 + 3\times 1\times 1000 = 3048 $
$\large III^{rd}\space part$
In previous part $M[2,4]$ is minimum
$M[1,4] = M[1,1] + M[2,4] + P_0P_1P_4 = 123 + 2\times 25\times 1=173 $
$M[2,4]$ is minimum so
$M[2,5] = M[2,4] + M[5,5] + P_1P_4P_5 = 123 + 25\times 1\times 1000= 25123$
$\large IV^{th}\space part$
$M[1,4]$ is minimum in previous part so
$M[1,5] = M[1,4] + M[5,5] + P_0P_4P_5 = 173 +2\times 1\times 1000 = 2173$
Now we can see-
$(F_1F_2F_3F_4F_5) \implies ((F_1F_2F_3F_4)F_5) \implies ((F_1(F_2F_3F_4))F_5) \implies((F_1(F_2(F_3F_4)))F_5) $
So Option C is right one