I have a proof for this, it was quite easy and straightforward, but that's the problem. It seemed too easy, I just want to make sure I didn't miss any important nuances or simplify things too greatly.
The full problem statement is as follows
If $\mathbb{F}$ is a field and $a \in \mathbb{F}$ is any element then prove that $\phi: \mathbb{F}(x) \to \mathbb{F}$ defined by $\phi(p(x)) = p(a)$ is a ring homomorphism from the ring $\mathbb{F}(x)$ to the ring $\mathbb{F}$.
Let $p(x), p'(x) \in \mathbb{F}(x)$ then we have $$\phi(p(x) + p'(x)) = (p(a) + p'(a)) = p(a) + p'(a) = \phi(p(x)) + \phi(p'(x))$$ $$\phi(p(x)p'(x)) = (p(a)p'(a)) = p(a)p'(a) = \phi(p(x))\phi(p'(x))$$
Essentially since $\mathbb{F}(x)$ and $\mathbb{F}$ are rings where multiplication is defined? Is it that straightforward?
In the terms of your problem statement, it probably is that simple and is OK.
However IMHO it is worthwhile making a distinction between $p$, a polynomial, and $p(x)$, which is a value of a polynomial. Then the problem would define the homomorphism by $$\phi(p)=p(a)$$ and the proof would be $$\phi(p+p')=(p+p')(a)=p(a)+p'(a)=\phi(p)+\phi(p')\ .$$ Depending on the approach you have taken, the reason for the middle step would either be a previously proved theorem, or the definition of polynomial addition.