Knowing that $$\tag{1} L= -mc\sqrt{-\eta_{ab}\frac{d\xi^a}{d\lambda}\frac{d\xi^b}{d\lambda}}$$ we get
$$\tag{2} p_a=\frac{\partial L}{\partial(d\xi^a/d\lambda)} = m\eta_{ab}u^b.$$
How come? If I differentiated this $L$ with respect to
$$\tag{3} d\xi^a/d\lambda$$
I get whole different answer. Shouldn't it be
$$\tag{4} p_a= (-mc)\left(-\eta_{cd}\frac{d\xi^c}{d\lambda}\frac{d\xi^d}{d\lambda}\right)^{-1/2}\left( -\eta_{ab}\frac{d\xi^b}{d\lambda} \right) ~?$$
How was this performed?
Your last expression (4) is equal to (2), you just have to realize what does it say. $\lambda$ isn't $\tau$ and $$u^c = \frac{d\xi^c}{d \tau} = \frac{d\xi^c}{d \lambda} \frac{d\lambda}{d \tau}$$ If you look back to your Lagrangian and how it was derived, you should be able to say what is $d\lambda/d\tau$.
To be very explicit, the action of a free relativistic particle is $$ S = -mc^2\int d\tau = -mc^2\int \frac{d\tau}{d\lambda} d\lambda $$ where we have used a reparametrization into a general parameter $\lambda$. Now you have $$L = -mc^2 \frac{d \tau}{d \lambda}$$ When you compare this with your equation (1) and with the knowledge $d\lambda/d\tau= (d\tau/d\lambda)^{-1}$, you should get the expression for $u^c$ in terms of $\lambda$-parametrization very easily. When you put this expression into (2), you will get (4).
