Let us build $E'$, the symmetrical of $E$ with respect to $P$. If $E$ is latticeal, then $E'$ is latticeal, and this implies $E' \notin \text{int}(ABC)$. We deduce that $E' = A$, otherwise $A \in (E'P)$ and through a translation we get that there exists a latticeal point lying on $PE$, contradiction. So in this case $\frac{AP}{PE} = 1$.
If $E$ is not a latticeal point, and on ($BC$) lies some other latticeal point $Q$, since $Q \neq E$ we can restrict the problem to either triangle $ABQ$ or $ACQ$, with the same initial hypothesis. Thus we can assume that the only latticeal points on the sides of the triangle are on $[AB]$ and $[AC]$.
We shall prove that, over all configurations,
$$\max \frac{AP}{PE} = 5,$$
value which can actually be reached by a proper configuration. Rather than looking for $\max \frac{AP}{PE}$, we will establish $\min \frac{PE}{AE} = \min \frac{[BPC]}{[ABC]}$ (where by $[XYZ]$ we mean the area of $\triangle XYZ$).
Let us notice that for the triangle $BPC$, the only latticeal points that are lying either on its sides, or in its interior, are just its vertices. We deduce from Pick's Theorem that $[BPC] = \frac{1}{2}$. Let us denote by $\beta$ and $\gamma$ the number of latticeal points on the sides ($AB$), respectively ($AC$). Availing ourselves again of Pick's Theorem, we find that $[ABC] = \frac{\beta + \gamma + 3}{2}$.
We get that $\frac{PE}{AE} = \frac{1}{\beta + \gamma + 3}$. We want to prove that $\frac{PE}{AE} \ge \frac{1}{6}$. We contrariwise assume $\frac{PE}{AE} < \frac{1}{6}$, and thus $\beta + \gamma \ge 4$.

Denote by $B_0, B_1, \dots, B_{\beta+1}$ the lattice points on $[AB]$, with $B_0 = A$, $B_{\beta+1} = B$, and by $C_0, C_1, \dots, C_{\gamma+1}$ the lattice points on $[AC]$, with $C_0 = A$, $C_{\gamma+1} = C$.
Let us consider the triangles of the type $PC_iC_{i+1}$ with $i = 0, \dots, \gamma$. They share the property that no lattice points lie inside or on the sides, other than the vertices, so $[PC_iC_{i+1}] = \frac{1}{2}$. It follows that $AC_1 = C_1C_2 = \dots = C_\gamma C$. With a similar reasoning we have also $AB_1 = B_1B_2 = \dots = B_\beta B$. Thus $\frac{AC_j}{C_jC_{j+1}} = \frac{AB_j}{B_jB_{j+1}}$, and so $C_iB_i \parallel C_{i+1}B_{i+1}$, for all $i \in \{1, 2, \dots, \min\{\beta, \gamma\}\}$.
We also deduce that on the segment $(B_iC_i)$ lie $i-1$ lattice points. This is because we can construct $i-1$ parallelograms with vertices of the type $B_lC_lB_iD$ with $l < i$, or of the type $B_lC_lC_iD$ with $D \in (B_iC_i)$. From effectively writing the relationships between coordinates in a parallelogram, we get that $D$ is a lattice point. Thus if $\min\{\beta, \gamma\} \ge 2$, the segments $(B_i, C_i)$ would contain at least three lattice points (the ones from $(B_2C_2)$ and $(B_3C_3)$), which is obviously false since $P$ is on at most one of the segments $(B_iC_i)$. This means $\min\{\beta, \gamma\} \le 1$, with say $\beta \ge \gamma$.

First case. $\gamma = 1$. From $\beta + \gamma \ge 4$ follows that $\beta \ge 3$ and that $C_1$ is the midpoint of $AC$. Now since $B_1$ is the midpoint of $AB_2$ we have that if we take $M$ the midpoint of $CB_2$, we get the parallelogram $B_2B_1C_1M$, and since all the points $B_2, B_1, C_1$ are lattice, we get $M$ lattice. Since $M$ is inside the triangle $ABC$, it must be $P$. The quadrilateral $BB_\beta PC$ is convex since $B_\beta \in (B_2B)$. We note that the triangles $BB_\beta C$ and $BB_\beta P$ have the property that there are no lattice points inside or on their sides, except their vertices, so $[BB_\beta P] = [BB_\beta C] = \frac{1}{2}$, hence $BB_\beta \parallel PC$, which is false.

Second case. $\gamma = 0$. It follows that $\beta \ge 4$.

LEMMA. Let $XYZT$ be a convex quadrilateral with lattice vertices, with the property that no lattice points lie in its interior, nor on its sides, except for its vertices. Then $XYZT$ is a parallelogram.
Proof. From Pick's theorem we know that $[XYZ] = [XYT] = \frac{1}{2}$, and so $XY \parallel ZT$. Similarly $XT \parallel YZ$, which ends the proof. $\square$

Let us consider the quadrilateral $AB_1PC$, with $AB_1 \cap PC \ne \emptyset$. If it is convex, the previous LEMMA implies a contradiction. Thus $[AB_1] \cap PC \ne \emptyset$. Also consider the quadrilateral $BB_\beta PC$. By the same reasoning, we must have $[BB_\beta] \cap PC \ne \emptyset$. This is the desired contradiction and ends the solution of the problem.