To find the sum of the given arithmetic series, we need to determine the number of terms and the common difference.

The first term, a, is 0, and the common difference, d, is -2 (since we are subtracting 2 from each term).

We can use the formula for the sum of an arithmetic series:

Sn = (n/2)(2a + (n-1)d)

where Sn is the sum of the series, n is the number of terms, a is the first term, and d is the common difference.

Plugging in the values, we have:

Sn = (n/2)(2(0) + (n-1)(-2))
Sn = (n/2)(-2(n-1))
Sn = -n(n-1)

Now we need to find the value of n. The last term of the series is 2022, so we can set -n(n-1) equal to 2022 and solve for n:

-n(n-1) = 2022
n^2 - n = -2022
n^2 - n + 2022 = 0

Using the quadratic formula, we find:

n = (-(-1) ± √((-1)^2 - 4(1)(2022))) / (2(1))
n = (1 ± √(1 + 8088)) / 2
n = (1 ± √8089) / 2

Since n must be a positive integer, we take the positive square root:

n = (1 + √8089) / 2

Now we can substitute this value of n back into the formula for Sn to find the sum:

Sn = -n(n-1)
Sn = -((1 + √8089) / 2)((1 + √8089) / 2 - 1)
Sn = -((1 + √8089) / 2)(-√8089 / 2)
Sn = ((1 + √8089) / 2)(√8089 / 2)
Sn = (1/4)(√8089 + 8089)

Therefore, the sum of the series P = 0 - 2 + 4 - 6 + ... + 2020 - 2022 is (1/4)(√8089 + 8089).