2017 is a prime

Since $1999$ is prime, for $A$ , $B$ or $C$ to be perfect squares, they need to have it with an even exponent in their prime factorization, but

$1997\cdot 1995\cdot 1993+16\equiv (-2)(-4)(-6)+16\equiv -32 \mod 1999$

$\Longrightarrow A$ is not the answer

$\begin{cases} \gcd(1997,1999)=1\\ 1995\cdot 1993+16\equiv (-4)(-6)+16\equiv 40 \mod 1999 \end{cases}$

$\Longrightarrow B$ is not the answer

$\begin{cases} \gcd(1997,1999)=1\\ \gcd(1995,1999)=1\\ 1993+16\equiv -6+16\equiv 10 \mod 1999 \end{cases}$

$\Longrightarrow C$ is not the answer

For $D$ I'll consider a general case

$\begin{aligned} k(k-2)(k-4)(k-6)+16&=\color{#D61F06}k(k-6)\color{#3D99F6}(k-2)(k-4)\color{#333333}+16\\ &=\color{#D61F06}(k^2-6k)\color{#3D99F6}(k^2-6k+8)\color{#333333}+16\\ &=(k^2-6k)^2+8(k^2-6k)+16\\ &=(k^2-6k)^2+2\cdot 4(k^2-6k)+4^2\\ &=(k^2-6k+4)^2 \end{aligned}$

With $k=1999$ we get that

$D=1999\cdot 1997\cdot 1995\cdot 1993+16=(1999^2-6\cdot 1999+4)^2=3984011^2$

Is a perfect square. Hence $\boxed{D}$ is the answer