When the loudspeaker is arranged in a plane, I want to calculate the sound pressure level at a certain distance from the front side I would like to calculate the initial sound pressure that must be guaranteed to guarantee 100dB at a distance of 2000m with a frequency of 17kHz.
Next, I would like to know how to arrange a few loudspeakers with an output of 140 dB to ensure the initial sound pressure. Should all loudspeakers have the same phase? Which arrangement structure is better?
 A: As already posted in this answer, in order to calculate the sound pressure at a specific point in space, knowing the pressure in some other point, you can use the inverse square law (this is the simplest case). In order to do that you first have to specify what the second point is. You cannot calculate the sound pressure at the position of the source, since the law has a singularity at that point. On top of that it is not physically valid to calculate sound pressure for short distances from the source, since you will be in the near-field where the inverse square law does not hold. The law does hold for short distances only in the ideal case where the source has zero dimensions and so the near-field (which is frequency/wavelength and source dimension dependent) has also zero dimensions. Of course this is not the case in any real source.
Just for completeness, the inverse square law is stated as (in polar coordinates)
$$p(r) = p_{ref}(r_{ref}) \frac{1}{\frac{r}{r_{ref}}} \implies p(r) = p_{ref}(r_{ref}) \frac{r_{ref}}{r}$$
where $r$ is the distance from the source of interest and $r_{ref}$ is the reference distance, at which the pressure $p_{ref}$ is known. Note that when $r$ is less than $r_{ref}$ you get increase in pressure. Also note that there is no square in the law for pressure. This is due to the fact that the initial formulation of the law is for intensity and pressure is a derived quantity.
In addition to geometrical attenuation, you also have to take into account the air absorption which is frequency, temperature and humidity dependent. For this, you can have a look at the linked resources of the aforementioned answer to get some representative tabulated (or extract from plots) values of attenuation per distance for various frequencies and conditions.
Now, regarding the loudspeaker arrangement, according to references in this answer (this is just a reference but there's a plethora of other sources you can search for), the maximum increase in pressure (power/intensity) you can achieve by arranging many sources in a linear arrangement is when all of them have the same phase characteristics (thus achieving coherent summation) in the frequency of interest and are weighted with the same coefficients (effectively, having the same amplitude, which of course can be unity, or else, maximum gain). In this case you achieve an $N$ multiplicative gain from the array compared to the individual element, where $N$ is the number of elements.
Replicating the formula presented in the linked answer to make some notes.
$$Y(\psi) = \frac{1}{N} \frac{\sin \left( N \frac{\psi}{2}\right)}{\sin \left( \frac{\psi}{2}\right)}$$
where $\psi$ is given by $\psi = \frac{2 \pi}{\lambda} d \cos (\theta)$ with $\lambda$ being the wavelength corresponding to the frequency of interest, $d$ the interelement distance (here assumed to be uniform in the array, thus talking about ULAs - Uniform Linear Array -) and $\theta$ the angle of interest. You should look for $\theta = 90^{o}$ (broadside of the array) where the elements will be arranged to "shoot". This way you achieve maximum again, which as stated (and can also be verified with the formula) will be $N$ times that of an individual source. Needless to say that all sources are assumed to be identical.
In case you arrange the sources to "shoot" in the direction parallel to the array axis you will end up with an end-fire array where the treatment is somewhat different and you will have to introduce additional delays to each source in order to achieve coherent summation. I believe (haven't double checked that) that the gain will be the same in this case, which is again $N$ times the amplitude of individual sources.
Please keep in mind that in real life, if there is considerable distance between the elements, geometrical attenuation (inverse square) and air absorption may apply to each source and the above formula may not yield exact results.
