The system of nonlinear ODE is $$ \mathrm{i}\,s(\ddot p-\frac{1}{2}\sin{2p}\;\dot q^2)=\mathrm{i}\,c\sin p\;\dot q-a\sin p+b\cos p\cos q\,,\\ \mathrm{i}\,s(\sin^2p\;\ddot q+\sin{2p}\;\dot p\dot q)=-\mathrm{i}\,c\sin p\;\dot p-b\sin{p}\sin{q}, $$ wherein $t\in[0,T],T>0$. Also, we have $a,b>0,c=1/2$. The positive $s$ must be small enough since the problem mathematically cares about the $s\rightarrow0$ limit. One can regard it as some strange motion on a sphere and $p,q$ as the two spherical coordinate angles. (So it would be very interesting to visualize any solution on the sphere.) Note that they can in general be complex (hence I said strange) because of the presence of $\mathrm{i}$, although set to be real at the boundaries.
One can scale the equations to have $a^2+b^2=1$. Then the interested domain of parameters is $c=1/2$, $s\ll1=(a^2+b^2)^{-1/2}\ll T$. For instance, let's take $T=6\sim20,s=0.03\sim0.08$ for a mild understanding of $\ll$ and the small $s$ limit. One can tune, of course.
The solution is to some extent expected to exhibit three regions in $t$. Near the left boundary $t-0 \sim O(s)$, rapidly varying from $(p,q)_{t=0}$ to some unknown $(\bar p_1,\bar q_1)$. Near the right boundary $T-t \sim O(s)$, rapidly varying from some unknown $(\bar p_2,\bar q_2)$ to $(p,q)_{t=T}$. And intermediately, slowly varying from $(\bar p_1,\bar q_2)$ to $(\bar p_2,\bar q_2)$. It's possible that $(\bar p_1,\bar q_1)=(\bar p_2,\bar q_2)$ in some cases. Usually, there are more than one valid solutions.
For instance, we hope to solve the following two representative cases $$ a=1,b=0,p(0)=p(T)=2\pi/3,q(0)=q(T)=\pi/4,\\ a=1/2,b=\sqrt{3}/2,p(0)=p(T)=3\pi/4,q(0)=\pi/4,q(T)=3\pi/4. $$ I tried NDSolve as follows without any success.
max = 10; s = 0.06; a = 1; b = 0; c = 0.5;
s = NDSolve[{I s (p''[t] - 1/2 Sin[2 p[t]] (q'[t])^2) ==
I c Sin[p[t]] q'[t] - a Sin[p[t]] + b Cos[p[t]] Cos[q[t]],
I s (Sin[p[t]]^2 q''[t] + Sin[2 p[t]] p'[t] q'[t]) == -I c Sin[
p[t]] p'[t] - b Sin[p[t]] Sin[q[t]], p[0] == 2 \[Pi]/3,
p[max] == 2 \[Pi]/3, q[0] == \[Pi]/4, q[max] == \[Pi]/4}, {p,
q}, {t, 0, max}]
A more demanding question
Ultimately, I hope to solve the case with following additional integral terms (setting, e.g., all $K_i=0.1,0.01$) added to the right hand side of the above two equations, respectively. It briefly damps the above motion. Either the case with or without integrals is interested in. If you can solve both, certainly wonderful. But solving only the above pure ODE is good enough for posting an answer. I just feel that it mightn't be good to raise a separate question for the integral case. $$ \int_{0}^T{\frac{\mathrm{d}t'}{(t-t')^2}[K_1\cos{p_t} \cos{q_{t}}\sin{p_{t'}} \cos{q_{t'}}+K_2\cos{p_t} \sin{q_{t}}\sin{p_{t'}} \sin{q_{t'}}-K_3\sin{p_t} \cos{p_{t'}}]}\\ \int_{0}^T{\frac{\mathrm{d}t'}{(t-t')^2}[-K_1\sin{p_t} \sin{q_{t}}\sin{p_{t'}} \cos{q_{t'}}+K_2\sin{p_t} \cos{q_{t}}\sin{p_{t'}} \sin{q_{t'}}]} $$
-b Cos[p[t]] Cos[q[t]] + a Sin[p[t]]==0andb Sin[p[t]] Sin[q[t]]==0. Solve this for the intialt =0 p[t]->2 Pi/3`` and q[t]->Pi/4to get `{{a -> 0, b -> 0}} $\endgroup$