If $X$ is a noetherian separated scheme and $X_{red}$ its reduction , we have $K_0(X)=K_o(X_{red})$: in other words $K_o$ doesn't see nilpotents .
Much more generally and profoundly, Quillen has proved that for all his $K$-theory groups, $K_i(X)=K_i(X_{red})$.
In your particular case you thus have $$K_0(\mathbb P^n_D)=K_0(\mathbb P^n_k)=\mathbb Z[T]/(T^{n+1})$$

As for $K^0$, a special case of a theorem of Berthelot (SGA 6, Exposé VI, Théorème 1.1, page 365) states that, for any commutative ring $A$, we have $K^0(\mathbb P^n_A)=K^o(A)[T]/(T^{n+1})$, where $T$ is an indeterminate.
If $A=D=k[\epsilon]$, we have $K^0(D)=\mathbb Z$, since projective modules over local rings (like $D$) are free.
So here too $$K^0(\mathbb P^n_D)=\mathbb Z[T]/(T^{n+1})$$

Bibliography
Srinivas has written this nice book on $K$-theory.

And as an homage to the recently sadly departed Daniel Quillen, let me refer to his groundbreaking paper
"Higher algebraic $K$-theory I", published in Springer's Lecture Notes LNM 341.

If $X$ is a noetherian separated scheme and $X_{red}$ its reduction , we have $K_0(X)=K_o(X_{red})$: in other words $K_o$ doesn't see nilpotents .
Much more generally and profoundly, Quillen has proved that for all his $K$-theory groups, $K_i(X)=K_i(X_{red})$.
In your particular case you thus have (in the following $T$ is an indeterminate) $$K_0(\mathbb P^n_D)=K_0(\mathbb P^n_k)=\mathbb Z[T]/(T^{n+1})$$

As for $K^0$, a special case of a theorem of Berthelot (SGA 6, Exposé VI, Théorème 1.1, page 365) states that, for any commutative ring $A$, we have $K^0(\mathbb P^n_A)=K^o(A)[T]/(T^{n+1})$.
If $A=D=k[\epsilon]$, we have $K^0(D)=\mathbb Z$, since projective modules over local rings (like $D$) are free.
So here too $$K^0(\mathbb P^n_D)=\mathbb Z[T]/(T^{n+1})$$

Bibliography
Srinivas has written this nice book on $K$-theory.

And as an homage to the recently sadly departed Daniel Quillen, let me refer to his groundbreaking paper
"Higher algebraic $K$-theory I", published in Springer's Lecture Notes LNM 341.