jmc

Note that $p(x)$ has degree at most 2. Also, $p(a)=p(b)=p(c)=1.$ Thus, the polynomials $p(x)$ and 1 agree at three different values, so by the Identity Theorem, they are the same polynomial. Hence, the degree of $p(x)$ (which is the constant polynomial 1) is $\boxed{0}.$

You can check manually that $$p(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-a)(x-c)}{(b-a)(b-c)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$$simplifies to 1.