Let $ q(x) = p(x) - x $. Then $ q(x) $ has degree 100, and $ q(1) = q(2) = \dots = q(100) = 0, $ so \[q(x) = c(x - 1)(x - 2) \dotsm (x - 100)\]for some constant $ c $. Since $ p(101) = 102, $ $ q(101) = 1 $. Setting $ x = 101 $ in the equation above, we get \[q(101) = 100! \cdot c,\]so $ c = \frac{1}{100!} $. Then \[q(x) = \frac{(x - 1)(x - 2) \dotsm (x - 100)}{100!}.\]In particular, \[q(102) = \frac{101 \cdot 100 \dotsm 2}{100!} = 101,\]so $ p(102) = q(102) + 102 = 101 + 102 = \boxed{203} $.