a. We will show that for all $k > 0$, the number $n = (p k + 1)(p - 1)$ satisfies $p \mid n \cdot 2^{n} + 1$. Indeed, by Fermat's Little Theorem, we have
$$2^{p-1} \equiv 1 \pmod{p} \text{ so } 2^{n} = 2^{(p k + 1)(p - 1)} \equiv 1 \pmod{p}.$$
And then
$$n \cdot 2^{n} + 1 \equiv (p k + 1)(p - 1) \cdot 1 + 1 = p^{2} k - p k + p \equiv 0 \pmod{p}.$$
Since there are infinitely many numbers of the form $(p k + 1)(p - 1)$, we get the conclusion.

b. Notice that $n$ is periodic modulo $3$, with a period of $3$, and $2^{n}$ is periodic modulo $3$, with period $2$. Hence, $n \cdot 2^{n} + 1$ is periodic, with period (at most) $6$, and only the first $6$ positive integers need to be analyzed.
The answer is $n = 6k + 1$ or $n = 6k + 2$.