Video solution:

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Written solution:

Because $BC=CD$, the median from $C$ to $BD$ is perpendicular to $BD$. Thus $\triangle BPC$ is a right triangle. Let $\angle DBC=\alpha$. Since $AB\parallel CD$, we also have $\angle ABD=\alpha$, so $\triangle BPC\sim\triangle BDA$.

Since $P$ is the midpoint of $BD$, we have $BD/BP=2$. In the similarity, $BC$ corresponds to $AB$, so

$$\frac{AB}{BC}=2,\qquad AB=2\cdot43=86.$$

Also, $\triangle ABO\sim\triangle CDO$, so

$$\frac{BO}{OD}=\frac{AB}{CD}=2.$$

Since $OP=11$ and $P$ is the midpoint of $BD$, write $BP=PD=t$. Then $BO=t+11$ and $OD=t-11$, so

$$\frac{t+11}{t-11}=2.$$

This gives $t=33$, hence $BD=66$. Finally, $\triangle ABD$ is right, so

$$AD=\sqrt{AB^2-BD^2}=\sqrt{86^2-66^2}=4\sqrt{190}.$$

Thus $m+n=4+190=194$.

Thus, D is the correct answer.