jmc

Let $q(x)=p(x)-x.$ Then $q(x)$ has degree 100, and $q(1)=q(2)=\cdots =q(100)=0,$ so $$q(x)=c(x-1)(x-2)\cdots (x-100)$$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 $$q(x)=\frac{(x-1)(x-2)\cdots (x-100)}{100!}.$$In particular, $$q(102)=\frac{101\cdot 100\cdots 2}{100!}=101,$$so $p(102)=q(102)+102=101+102=\boxed{203}.$