Somebody stole the constant!

For the sake of variety, let me outline the approach I took.

Setting the derivative equal to 0, we see that $P(x)$ has critical points at $x=1$ and at $x=13$ , regardless of the value of $a$ . To get a double root, we want $P(1)=19+a=0$ or $P(13)=-845+a=0$ , so that the possible values of $a$ are -19 and 845, with their sum being $\boxed{826}$

Let the roots of $P(x)$ be $r$ and $s$ , and let $r$ have multiplicity 2.

Then $P(x)=(x-r)^2(x-s)$

Expanding this yields $P(x)=x^3-(2r+s)x^2+(r^2+2rs)x-r^2s$

Now we can see that $21=2r+s$ , $39=r^2+2rs$ , and $a=-r^2s$ .

Solving the system of the first two equations yields two possible ordered pairs $(r,s)$ : $(1,19)$ and $(13,-5)$

We can plug those values into $a=-r^2s$ to get the two possible values for $a$ : $a=-19$ or $a=845$ . The sum of these possible values is $\boxed{826}$ .