Let's first understand and break down the given information. We are provided with the following details to solve for the value of $\mathbf{P}$:

1. Volume of acetic acid solution used: $100 \mathrm{~mL}$ of $0.5 \mathrm{M}$ acetic acid.

2. Volume of $\mathrm{NaOH}$ solution used for neutralization: $40 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{NaOH}$.

3. Surface area of the charcoal: $1.5 \times 10^2 \mathrm{~m}^2 \mathrm{~g}^{-1}$.

4. Avogadro's number: $\mathrm{N}_{\mathrm{A}}=6.0 \times 10^{23} \mathrm{~mol}^{-1}$.

Let's start with the concentration and volume of acetic acid used:

$$ \text{Moles of acetic acid initially} = 0.5 \text{ M} \times \frac{100 \text{ mL}}{1000} = 0.05 \text{ moles} $$

Next, let's determine the moles of $\mathrm{NaOH}$ used for neutralization:

$$ \text{Moles of } \mathrm{NaOH} = 1 \text{ M} \times \frac{40 \text{ mL}}{1000} = 0.04 \text{ moles} $$

Since $\mathrm{NaOH}$ completely neutralizes the unadsorbed acetic acid:

$$ \text{Moles of unadsorbed acetic acid} = 0.04 \text{ moles} $$

Therefore, the moles of acetic acid adsorbed on the charcoal is given by:

$$ \text{Moles of adsorbed acetic acid} = 0.05 - 0.04 = 0.01 \text{ moles} $$

Now, we need to calculate the total number of molecules of acetic acid adsorbed:

$$ \text{Number of molecules} = 0.01 \text{ moles} \times 6.0 \times 10^{23} \text{ molecules/mole} = 6.0 \times 10^{21} \text{ molecules} $$

Given the surface area of charcoal, we can find the total surface area:

$$ \text{Total surface area} = 1.5 \times 10^2 \text{ m}^2 / \text{g} \times 1 \text{ g} = 1.5 \times 10^2 \text{ m}^2 $$

Finally, we need to determine the area occupied by each molecule of acetic acid:

$$ \text{Area per molecule of acetic acid} = \frac{\text{Total surface area}}{\text{Number of molecules}} = \frac{1.5 \times 10^2 \text{ m}^2}{6.0 \times 10^{21} \text{ molecules}} $$

Let's perform the calculation:

$$ \text{Area per molecule of acetic acid} = \frac{1.5 \times 10^2}{6.0 \times 10^{21}} = 0.25 \times 10^{-19} \text{ m}^2 = 2.5 \times 10^{-20} \text{ m}^2 $$

Since the problem states that the area per molecule is $\mathbf{P} \times 10^{-23} \mathrm{~m}^2$, we equate:

$$ 2.5 \times 10^{-20} = \mathbf{P} \times 10^{-23} $$

Solving for $\mathbf{P}$:

$$ \mathbf{P} = \frac{2.5 \times 10^{-20}}{10^{-23}} = 2.5 \times 10^3 = 2500 $$

Therefore, the value of $\mathbf{P}$ is 2500.