Regarding the sum $\sum_{p \ \text{prime}} \sin p$ I'm very confident that 
$$\sum_{p \ \text{prime}} \sin p $$
diverges. Of course, it suffices to show that there are arbitrarily large primes which are not in the set $\bigcup_{n \geq 1} (\pi n - \epsilon, \pi n + \epsilon)$ for sufficiently small $\epsilon$. More strongly, it seems that $\sin p$ for prime $p$ is dense in $[-1,1]$. 
This problem doesn't seem that hard though. Here's something that (to me) seems harder. 

If $p_n$ is the nth prime, what is $$\limsup_{n \to +\infty} \sum_{p \ \text{prime} \leq p_n} \sin p?$$
  What is $$\sup_{n \in \mathbb{N}} \sum_{p \ \text{prime} \leq p_n} \sin p? $$

Of course, we can ask analogous questions for $\inf$. 
I'm happy with partial answers or ideas. For example, merely an upper bound.
 A: My answer here only includes partial results. First, we use Vinogradov's inequality:

Let $\alpha$ be a real number. If integers $a$ and $q$ satisfies $(a,q)=1$ and
  $$
\left| \alpha - \frac aq \right| \leq \frac 1{q^2},
$$
  then 
  $$
\sum_{n\leq N} \Lambda(n) e^{2\pi i \alpha n} = O\left( (Nq^{-1/2} +N^{4/5} + N^{1/2}q^{1/2} ) (\log N)^4 \right)
$$

With an error of $O(N^{1/2+\epsilon})$, we obtain the same upper bound for $\sum_{p\leq N} (\log p) \cdot e^{2\pi i \alpha p}$. 
Since we have finiteness of irrationality measure of $\pi$ : see this, we may use the continued fraction convergents $a/q$ for $\alpha = 1/(2\pi)$. Thus, it is possible to find a denominator $q$ of the continued fraction convergent of $1/(2\pi)$ such that $N^{1/7}<q<N^{99/100}$. Then Vinogradov inequality yields that there is $\delta>0$ such that 
$$
\sum_{p\leq N} (\log p)e^{i p} = O(N^{1-\delta}).
$$
Now, partial summation gives for some $\delta>0$, 
$$
\sum_{p\leq N} e^{ip} = O(N^{1-\delta}).
$$
Therefore, by taking imaginary parts, 
$$
\left|\sum_{p\leq N} \sin p \right| = O(N^{1-\delta}).
$$
With this result and partial summation, we obtain that 
$$
\sum_{p \ \mathrm{prime} } \frac{\sin p}p 
$$
converges. 
It will be possible to find the best $\delta>0$ in the upper bound:
$$
\left|\sum_{p\leq N} \sin p \right| = O(N^{1-\delta})
$$
by using Vinogradov's inequality more efficiently. 
