jmc

Since $PQ=2$ and $M$ is the midpoint of $PQ$, then $PM=MQ=\frac{1}{2}(2)=1$.

Since $△PQR$ is right-angled at $P$, then by the Pythagorean Theorem, $$RQ=\sqrt{P{Q}^2+P{R}^2}=\sqrt{2^2+(2\sqrt{3}{)}^2}=\sqrt{4+12}=\sqrt{16}=4.$$(Note that we could say that $△PQR$ is a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle, but we do not actually need this fact.)

Since $PL$ is an altitude, then $\angle PLR=90^{\circ }$, so $△RLP$ is similar to $△RPQ$ (these triangles have right angles at $L$ and $P$ respectively, and a common angle at $R$).

Therefore, $\frac{PL}{QP}=\frac{RP}{RQ}$ or $PL=\frac{(QP)(RP)}{RQ}=\frac{2(2\sqrt{3})}{4}=\sqrt{3}$.

Similarly, $\frac{RL}{RP}=\frac{RP}{RQ}$ so $RL=\frac{(RP)(RP)}{RQ}=\frac{(2\sqrt{3})(2\sqrt{3})}{4}=3$.

Therefore, $LQ=RQ-RL=4-3=1$ and $PF=PL-FL=\sqrt{3}-FL$.

So we need to determine the length of $FL$.

Drop a perpendicular from $M$ to $X$ on $RQ$.



Then $△MXQ$ is similar to $△PLQ$, since these triangles are each right-angled and they share a common angle at $Q$. Since $MQ=\frac{1}{2}PQ$, then the corresponding sides of $△MXQ$ are half as long as those of $△PLQ$.

Therefore, $QX=\frac{1}{2}QL=\frac{1}{2}(1)=\frac{1}{2}$ and $MX=\frac{1}{2}PL=\frac{1}{2}(\sqrt{3})=\frac{\sqrt{3}}{2}$.

Since $QX=\frac{1}{2}$, then $RX=RQ-QX=4-\frac{1}{2}=\frac{7}{2}$.

Now $△RLF$ is similar to $△RXM$ (they are each right-angled and share a common angle at $R$).

Therefore, $\frac{FL}{MX}=\frac{RL}{RX}$ so $FL=\frac{(MX)(RL)}{RX}=\frac{\frac{\sqrt{3}}{2}(3)}{\frac{7}{2}}=\frac{3\sqrt{3}}{7}$.

Thus, $PF=\sqrt{3}-\frac{3\sqrt{3}}{7}=\boxed{\frac{4\sqrt{3}}{7}}$.