A field extension of prime degree

Of course, any nonzero vector of a $1$-dimensional subspace generates that subspace.

Therefore in $F(a) = \{yd \mid d \in F\}$ you may as well choose $y=1\in F(a)$ as the generator. Thus $F(a) = \{d \mid d \in F\}=F$.

The general observation to be made is $[E:F]=1$ iff $E=F$. See if you can't convince yourself of this.

You will move ahead by applying this to both $[F(a):F]$ and $[E:F(a)]$