= Solution
Put $u=t-p$ and write $a(t)=u^r b(u)$, where $b(0)\ne0$. The unit $b$ has a square root in $\mathbb C[[u]]$. Dividing the radical by that square root and by $u^{\lfloor r/2\rfloor}$ reduces the completed extension to
$$
\mathbb C((u))(\sqrt{u^{\,r\bmod2}}).
$$
If $r$ is even, it is unramified and locally splits; if $r$ is odd, adjoining $u^{1/2}$ gives <ramification index> two. At points where $a(p)\ne0$, $r=0$ and there is no branch. Therefore
$$
\boxed{p\in\mathbb C\text{ branches exactly when it is a zero of odd multiplicity.}}
$$
For infinity set $u=1/t$. If $D=\deg a$,
$$
a(1/u)=u^{-D}b(u),\qquad b(0)\ne0.
$$
The same <quadratic branch parity criterion> applies to the negative <valuation> $-D$, so
$$
\boxed{\infty\text{ branches exactly when }\deg a\text{ is odd}.}
$$
The generic irreducibility excludes $a$ being a square. Local splitting at the even-order places is compatible with that global irreducibility. The number of odd finite orders plus the possible odd infinity order is even, as also follows from the sum of orders of a rational function being zero.
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