= Solution
Put $k=p+1$, $\mu=GM_\star$ and $n_{\rm pl}=\sqrt{\mu/a_{\rm pl}^3}$. To leading order the <mean-motion resonance> fixes $a=a_{\rm pl}k^{-2/3}$. The <specific angular momentum> is
$$
h^2=\mu q(1+e)\simeq2\mu q.
$$
The <asteroid>'s instantaneous <angular speed> is $h/r^2$. Thus the <high-eccentricity angular-speed crossover> satisfies $h/r_x^2=n_{\rm pl}$, giving
$$
\boxed{r_x=a_{\rm pl}\left(\frac{2q}{a_{\rm pl}}\right)^{1/4}},\qquad \dot\theta<n_{\rm pl}\quad\hbox{for }r>r_x.
$$
Here $q\ll r_x\ll a$, so a <parabolic Kepler orbit> gives the local motion accurately. Its polar equation is $r=2q/(1+\cos f)$. Write $f_x=\pi-\epsilon_x$. Since $1+\cos f_x\simeq\epsilon_x^2/2$,
$$
\boxed{f_x\simeq\pi-2\sqrt{q/r_x}=\pi-2^{7/8}(q/a_{\rm pl})^{3/8}}.
$$
This is the outward crossing; the inward crossing has <true anomaly> $-f_x$ modulo $2\pi$.
For the <pericentre-to-crossover flight time> use $D=\tan(f/2)$. The <parabolic Kepler orbit> has $r=q(1+D^2)$ and $dt=r^2\,df/h$, whence direct integration gives the <Barker equation>
$$
t(f)-t(0)=\sqrt{\frac{2q^3}{\mu}}\left(D+\frac{D^3}{3}\right).
$$
At the crossover $D_x\simeq\sqrt{r_x/q}\gg1$. Consequently the <planet>'s angular displacement is
$$
n_{\rm pl}t_x\simeq\frac{\sqrt2}{3}\left(\frac{r_x}{a_{\rm pl}}\right)^{3/2}
=\boxed{\frac{2^{7/8}}3\left(\frac{q}{a_{\rm pl}}\right)^{3/8}}.
$$
The error tends to zero as $q/a_{\rm pl}\to0$ at fixed $k$; it includes the finite binding energy of the elliptic <Kepler orbit> as well as the subleading term in the <Barker equation>.
At the $j$th <pericentre> passage the <asteroid>'s unwrapped <mean longitude> is $\lambda=\varpi+2\pi j$. The <resonant argument> therefore gives
$$
\lambda_{\rm pl}-\varpi=\frac{\phi+2\pi j}{k}\pmod {2\pi}.
$$
Thus $\phi/k$ specifies the <planet>'s direction relative to the <asteroid>'s <longitude of pericentre> when the <asteroid> passes <pericentre>, with $k$ branches separated by $2\pi/k$. It is not the instantaneous true-longitude separation throughout the orbit. At high <orbital eccentricity> the <asteroid> quickly sweeps to a direction nearly opposite <pericentre>, while the <planet> scarcely moves. During the long outer excursion the <asteroid>'s direction changes slowly and the <planet> advances substantially. This makes an <astronomical conjunction> during the outer excursion much more dangerous than an <astronomical conjunction> close to <pericentre>.
At the following <apocentre>, the <planet> has advanced $\pi/k$. The <asteroid>'s direction in the <rotating reference frame> is consequently
$$
\psi_{{\rm apo},j}=\pi-\frac{\phi+(2j+1)\pi}{k}.
$$
These $k$ <apocentre> directions are equally spaced. <Phase protection of interior integer resonances> places the <planet> in the largest gap between them, rather than at an <apocentre> direction. For $k=2$, $\phi=\pi$ puts an <apocentre> toward the <planet>, whereas $\phi=0$ puts the <apocentres> at $\pm\pi/2$. For $k=3$, $\phi=0$ puts an <apocentre> toward the <planet>, whereas $\phi=\pi$ puts the <apocentres> at $\pi/3,\pi,5\pi/3$. Hence the geometrically favored <resonant-argument libration> centres in this regime of high <orbital eccentricity> are
$$
\boxed{\phi\simeq0\quad(2:1),\qquad \phi\simeq\pi\quad(3:1)}\pmod {2\pi}.
$$
This is an encounter-avoidance argument for likely stability, not a calculation of a resonant <Hamiltonian> or a claim that every orbit near either centre is stable at arbitrary <orbital eccentricity>.
The following rotating-frame drawing uses $a_{\rm pl}=1$, $a=3^{-2/3}$, $q=0.002$ and $\phi=\pi$. It follows three successive <Kepler orbits>, which fill one <planet> <orbital period> and close the pattern. The blue segment is just the first outward half-orbit. The marked crossover is a maximum of its rotating-frame polar angle, since $d\psi/dt=h/r^2-n_{\rm pl}$ changes from positive to negative there. The <planet> stays at $(1,0)$; the three <apocentres> avoid that direction.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-64-resonant-orbit.png]
{title=Phase-protected high-eccentricity 3:1 orbit in the planet's rotating frame}
To estimate <angular-momentum kicks to nearly radial orbits>, note that $q\simeq h^2/(2\mu)$, so an order-one fractional change in $q$ requires $|\Delta h|$ of order $h$. A characteristic distant encounter at radii of order $a_{\rm pl}$ has tangential acceleration of order $GM_{\rm pl}/a_{\rm pl}^2$ and duration of order $n_{\rm pl}^{-1}$. Its <torque> per unit <asteroid> <mass> therefore produces
$$
|\delta h|\sim\frac{GM_{\rm pl}}{a_{\rm pl}n_{\rm pl}}
=\frac{M_{\rm pl}}{M_\star}\sqrt{GM_\star a_{\rm pl}}.
$$
With the most favorable coherent signs, the crude encounter count is
$$
\boxed{N_{\rm coherent}\sim\max\left[1,\frac{M_\star}{M_{\rm pl}}\sqrt{\frac{q}{a_{\rm pl}}}\right]}.
$$
Order-one factors depend on $p$ and encounter geometry. Uncorrelated signs instead give a <random walk> count of order $(M_\star/M_{\rm pl})^2q/a_{\rm pl}$ when this is large. A librating phase-protected orbit can suppress the kicks even further.
There is an important limit to calling this a minimum. The question does not specify an encounter <impact parameter>. For a weak close encounter with <impact parameter> $b$ and relative speed $u\sim\sqrt{GM_\star/a_{\rm pl}}$, the <impulse> estimate gives $|\delta h|\sim GM_{\rm pl}a_{\rm pl}/(bu)$ and hence $N\sim(M_\star/M_{\rm pl})(b/a_{\rm pl})\sqrt{q/a_{\rm pl}}$, capped below by one. In the 2:1 case the <Kepler orbit> can cross the <planet>'s orbit, so one suitably close encounter can suffice: there is no impact-parameter-independent large lower bound. For $k\ge3$, $Q\simeq2a_{\rm pl}k^{-2/3}<a_{\rm pl}$ bounds the separation below by $a_{\rm pl}-Q$, and supplies a geometric factor at fixed $k$. The boxed count is the usual orbital-scale, maximally coherent estimate; the assumptions are essential.
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