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If A and C are disjoint, K =1 suffices, and this is provable by induction on the number of segments comprising A and B. If A and C are not disjoint, let D denote the intersection. Let A=A1*DA2 and C=C1*DC2. The diameter of the planar projection of Ai is at most that of Ci. Hence K=3 is an upper bound on the ratio of diameters of the planar projections of A and C.
It is more straightforward to see K=3 is a universal upper bound. Let $Q$ denote the universal cover of $P$. Note $Q$ is CAT(0). Take any two geodesics segments A and B in $Q$. In $Q$, naturally project A into B. Call the image C. Now map A and C naturally into the plane and compare the ratios of the their diameters, using absolute value.
There is a genuine gap since K=1.5 can actually happen. Place the letter V in the upper half plane with vertex at (0,0) and angles 60 degrees with w.r.t. x axis. Now place a different V with origin vertex, with angles 30 degrees and very long edges. Now project orthogonally the first V into the 2nd V while fixing the origin. This is purportedly the worst case, yielding a kind of poor man's Gehring-Hayman Theorem.
The relevance of the question is a `yes' answer would characterize T_0 qcb spaces (quotients of 2nd countable spaces), as quotients of separable metric spaces, since sequential spaces are precisely the quotients of a metrizable spaces, and since 2nd countable spaces are sequential, as noted in the question. Basic properties of qcb spaces are not obvious: mathematik.tu-darmstadt.de/~streicher/GrSt.pdf.
