I believe option number 2 is the correct one, given the circumstances that you describe.
At its base, you want each question on this multiple-choice test, regardless of how many possible answers it has, to be worth $1$ point if the student has demonstrated he knows the answer to the question, and $0$ points otherwise. This is the only fundamental way to make each question "worth" the same. Anything else is just clever rescaling.
The most common way to formulate this is to give $1$ point for a correct answer, and $0$ points for an incorrect answer. However, complicating this is the fact that a student might get the correct answer purely through guessing. So, you decide to give them an "incentive" (really just removing the edge from the former choice) to not answer if they have no idea, by giving them $1/n$ points, the expected value from guessing randomly, for leaving the question blank. In your scenario, this is $1/4$ point for a question with four choices, and $1/2$ point for a true-false question.
If you scale this up by a factor of $4$ to make the point values all integers, you get:
$$\begin{matrix}
P_4 = 4 & O_4 = 1 & p_4 = 0\\
P_2 = 4 & O_2 = 2 & p_2 = 0
\end{matrix}$$
And if you subtract enough from each row to make $O_n$ equal to $0$ in all cases, you get:
$$\begin{matrix}
P_4 = 3 & O_4 = 0 & p_4 = -1\\
P_2 = 2 & O_2 = 0 & p_2 = -2
\end{matrix}$$
which is exactly your second option.
Another, less common way to formulate this is to give $1$ point for a correct answer, and $0$ points to the expected value of random guessing. This also fits the description, assuming that you equate random guessing to "not knowing anything about the question". If this is the case, then you'll want a matrix where the following is true:
$$\begin{matrix}
P_4 = 1 & O_4 = 0 & 3p_4 + P_4 = 0\\
P_2 = 1 & O_2 = 0 & p_2 + P_2 = 0
\end{matrix}$$
Solve those two equations on the right and we get:
$$\begin{matrix}
P_4 = 1 & O_4 = 0 & p_4 = -1/3\\
P_2 = 1 & O_2 = 0 & p_2 = -1
\end{matrix}$$
And scale it up by three to make the point values integers:
$$\begin{matrix}
P_4 = 3 & O_4 = 0 & p_4 = -1\\
P_2 = 3 & O_2 = 0 & p_2 = -3
\end{matrix}$$
and we get your third option.
So we've narrowed our choices down to your options 2 and 3 because the first two don't account for the model of "$1$ point for knowing, $0$ points for not knowing".
At this point, I'd go with option 2 for two reasons:
It follows the more conventional formulation of $1$ point for a correct answer and $0$ points for an incorrect answer.
It feels more aesthetically pleasing. Imposing a penalty of $-3$ points for an incorrect answer to a true-and-false feels like way too much gravity to put onto a true-false question.
Note that the AP exams, before they dropped penalties altogether, went with option 3, taking away $1/4$ point for incorrect answers to questions with $5$ choices and $1/3$ point for incorrect answers to question with $4$ choices. However, there's not much difference between $1/4$ and $1/3$, while there is a huge difference between $-1/3$ and $-1$.
That being said, I really do need to criticize the usage of penalties for incorrect answers on multiple-choice exams here. I completely fail to understand how getting an answer incorrect somehow demonstrates that the student knows less than nothing about the question, or conversely how leaving the answer blank means that the student knows any more about the question than how to take the test.
You may be (and most people are) doing this to compensate for the fact that getting an answer correct does not necessarily demonstrate that the student knows the answer to the question. But that is a flaw inherent to multiple-choice tests. You can't get rid of that just by making leaving an answer blank an equal alternative (which is what the expected-value penalty really does).