The sum of the first ten terms of an A.P. is $160$ and the sum of the first two terms of a G.P. is $8$. If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to common difference of the A.P., then the sum of all possible values of the first term of the G.P. is:

Let the first term of the A.P. be $a$ and its common difference be $d$.
Let the first term of the G.P. be $A$ and its common ratio be $R$.

Given that the sum of the first ten terms of the A.P. is $160$:
$\dfrac{10}{2} [2a + (10 - 1)d] = 160$
$5(2a + 9d) = 160$
$2a + 9d = 32$

Given that the sum of the first two terms of the G.P. is $8$:
$A + AR = 8$
$A(1 + R) = 8$

We are also given that $a = R$ and $A = d$. Substituting these into the first equation:
$2R + 9A = 32$
$R = \dfrac{32 - 9A}{2}$

Substitute $R$ into the second equation:
$A \left(1 + \dfrac{32 - 9A}{2}\right) = 8$
$A \left(\dfrac{2 + 32 - 9A}{2}\right) = 8$
$A(34 - 9A) = 16$
$34A - 9A^2 = 16$
$9A^2 - 34A + 16 = 0$

This is a quadratic equation in $A$. The discriminant is $\Delta = (-34)^2 - 4(9)(16) = 1156 - 576 = 580 > 0$, so there are two distinct real values for $A$.

The sum of all possible values of the first term of the G.P. ($A$) is the sum of the roots of this quadratic equation:
Sum of roots $= -\dfrac{\text{coefficient of } A}{\text{coefficient of } A^2} = \dfrac{34}{9}$

Answer: $\dfrac{34}{9}$