I suppose you want $D$ to be a $K$-derivation, otherwise there are obvious counter-examples with $A=K$. Then the answer is yes, but for somewhat stupid reasons. First of all, $A$ is a product of a finite number of local rings, and $D$ respects this decomposition: if $e\in A$ is idempotent, from $e^2=e$ one gets $(2e-1)De=0$, hence $De=0$, and therefore $D(Ae)\subset Ae$. Thus one can assume that $A$ is local, with maximal ideal $\mathfrak{m}$. Let $x\in\mathfrak{m}$; there exists an integer $n\geq 1$ such that $x^n=0$ but $x^{n-1}\neq 0$. Then $nx^{n-1}D(x)=0$, which means that $D(x)$ cannot be invertible, hence $D(x)\in\mathfrak{m}$.

I suppose you want $D$ to be a $K$-derivation, otherwise there are obvious counter-examples with $A=K$. Then the answer is yes, but for somewhat stupid reasons. First of all, $A$ is a product of a finite number of local rings, and $D$ respects this decomposition: if $e\in A$ is idempotent, from $e^2=e$ one gets $(2e-1)De=0$, hence $De=0$, and therefore $D(Ae)\subset Ae$. Thus one can assume that $A$ is local, with maximal ideal $\mathfrak{m}$. Let $x\in\mathfrak{m}$; there exists an integer $n\geq 1$ such that $x^n=0$ but $x^{n-1}\neq 0$. Then $nx^{n-1}D(x)=0$, which means that $D(x)$ cannot be invertible, hence $D(x)\in\mathfrak{m}$.

Edit : As observed by @Keith Kearnes, the argument is completed as follows: we have shown $D(\mathfrak{m})\subset \mathfrak{m}$, so that $D$ induces a $K$-derivation of $A/\mathfrak{m}$. But $A/\mathfrak{m}$ is a finite extension of $K$, so any $K$-derivation of $A/\mathfrak{m}$ is zero. But this means $D(A)\subset \mathfrak{m}$.