= Solution
Factor the integrand in the <Schläfli contour integral for Legendre polynomials> as
$$
\frac{(z^2-1)^n}{(z-t)^{n+1}}
=\frac1{z-t}
\left(\frac{z^2-1}{z-t}\right)^n.
$$
Thus one may take
$$
f(z,t)=\frac1{z-t},
\qquad
\phi(z,t)=\log\left(\frac{z^2-1}{z-t}\right),
$$
with a branch of the logarithm chosen locally along the deformed contour. Differentiation gives
$$
\phi'(z,t)
=\frac{2z}{z^2-1}-\frac1{z-t}
=\frac{z^2-2tz+1}{(z^2-1)(z-t)}.
$$
For $t=\cos\theta$, its saddle points solve
$$
z^2-2(\cos\theta)z+1=0,
$$
and hence
$$
\boxed{z_\pm=\cos\theta\pm i\sin\theta=e^{\pm i\theta}}.
$$
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