1. **State the problem:** We have a circle with diameters \(\overline{AC}\) and \(\overline{BD}\) intersecting at center \(P\). The arcs \(BC\) and \(AD\) have measures \((4k + 159)^\circ\) and \((2k + 153)^\circ\) respectively. We need to find the measure of the major arc \(\overset{\frown}{BDC}\).\n\n2. **Understand the arcs:** Diameters divide the circle into 4 arcs: \(AB\), \(BC\), \(CD\), and \(DA\). Here, arcs \(BC\) and \(AD\) are given.\n\n3. **Sum of arcs in circle:** Total degrees in a circle is \(360^\circ\). So, \(AB + BC + CD + DA = 360^\circ\).\n\n4. **Opposite arcs of diameters are supplementary:** Since \(AC\) and \(BD\) are diameters, arcs \(AB + BC = 180^\circ\) and arcs \(CD + DA = 180^\circ\).\n\n5. **Express the sums:**\n- \(BC + AB = 180^\circ\)\n- \(CD + DA = 180^\circ\)\n\n6. **We know \(BC = 4k + 159\) and \(AD = 2k + 153\), so we want to find \(\overset{\frown}{BDC} = BD + DC = BA + AD + DC\). But since \(BD\) is diameter, major arc \(BDC = 360^\circ - \overset{\frown}{AC}\) (because \(AC\) is the other diameter).\n\n7. **Calculate \(\overset{\frown}{AC}\):** The arcs \(AB + BC = 180^\circ\) and \(CD + DA = 180^\circ\) so \(\overset{\frown}{AC} = AB + BC\) or equivalently \(AD + DC\). But clearer approach is to use given arcs:\nThe arcs around the circle are \(AB\), \(BC = 4k + 159\), \(CD\), \(DA = 2k+153\). Sums:\n- \(AB + BC = 180\)\n- \(CD + DA = 180\)\n\n8. **Find \(AB\) and \(CD\):**\n- \(AB = 180 - BC = 180 - (4k + 159) = 21 -4k\)\n- \(CD = 180 - DA = 180 - (2k + 153) = 27 - 2k\)\n\n9. **Sum of all arcs:**\n\(AB + BC + CD + DA = (21 - 4k) + (4k + 159) + (27 - 2k) + (2k + 153)\)\nSimplify:\n\(21 - 4k + 4k + 159 + 27 - 2k + 2k + 153 = 21 + 159 + 27 + 153 = 360\) degrees, which checks out.\n\n10. **Major arc \(\overset{\frown}{BDC}\) includes arcs \(BD + DC\), effectively \(BD\) diameter plus arc \(DC\). We can express it as:\n- \(\overset{\frown}{BDC} = 180^\circ + CD\)\n\nFrom step 8, \(CD = 27 - 2k\). So,\n\[\overset{\frown}{BDC} = 180 + (27 - 2k) = 207 - 2k.\]\n\n11. **Find \(k\) using angle sum in quadrilateral or arcs:** We can use the fact that arcs \(BC + AD\) are given. Since both pairs of arcs correspond to the circle divided by diameters, the setup suggests arcs are consistent.\nIf arcs \(BC\) and \(AD\) correspond to an endpoint, try summing:\n\((4k + 159) + (2k + 153) = 6k + 312\) which must be less than 360 since arcs do not overlap fully.\nBut to find \(k\) exactly, note that arcs \(\overset{\frown}{BC}\) and \(\overset{\frown}{AD}\) are opposite arcs along the diameters, so their sum must equal \(180^\circ\) because each pair of arcs defined by diameters sums to 180 degrees (since diameters split circle in halves). So\n\[ (4k + 159) + (2k + 153) = 180 \]\n\[6k + 312 = 180\]\n\[6k = 180 - 312 = -132\]\n\[k = -22\]\n\n12. **Calculate major arc measure with \(k = -22\):**\n\[\overset{\frown}{BDC} = 207 - 2(-22) = 207 + 44 = 251^\circ\]\n\n**Final answer:** The measure of the major arc \(\overset{\frown}{BDC} = 251^\circ\).